I think you'd benefit from loading the arrows.meta and bending libraries. UPDATE: I could not resist creating a style that allows one to put a bent arrow at an arbitrary position of the path.
\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{calc} % shapes,arrows,shapes.multipart,positioning, fit, were not used
\usetikzlibrary{arrows.meta,bending,decorations.markings,intersections} %< added
\tikzset{% inspired by https://tex.stackexchange.com/a/316050/121799
arc arrow/.style args={%
to pos #1 with length #2}{
decoration={
markings,
mark=at position 0 with {\pgfextra{%
\pgfmathsetmacro{\tmpArrowTime}{#2/(\pgfdecoratedpathlength)}
\xdef\tmpArrowTime{\tmpArrowTime}}},
mark=at position {#1-\tmpArrowTime} with {\coordinate(@1);},
mark=at position {#1-2*\tmpArrowTime/3} with {\coordinate(@2);},
mark=at position {#1-\tmpArrowTime/3} with {\coordinate(@3);},
mark=at position {#1} with {\coordinate(@4);
\draw[-{Stealth[length=#2,bend]}]
(@1) .. controls (@2) and (@3) .. (@4);},
},
postaction=decorate,
}
}
\begin{document}
\begin{tikzpicture}
%roads: grey graph
\begin{scope}[every node/.style={circle,thick,draw}]
\node (A) at (0,0) {A};
\node (B) at (3,0) {B};
\node (C) at (6,0) {C};
\node (D) at (3,3) {D};
\end{scope}
\begin{scope}[every edge/.style={draw=lightgray,line width=3pt}]
\path [-] (A) edge (B);
\path [-] (B) edge (C);
\path [-] (B) edge (D);
\end{scope}
%line graph
\begin{scope}[every node/.style={circle,thick,draw,scale=.7}]
\node (AB) at (1.5,-.5) {};
\node (BA) at (1.5,.5) {};
\node (BC) at (4.5,-.5) {};
\node (CB) at (4.5,.5) {};
\node (BD) at (3.5,1.5) {};
\node (DB) at (2.5,1.5) {};
\end{scope}
\begin{scope}[line width=1pt]
% bottom left
\draw[arc arrow=to pos 0.55 with length 3mm] (AB) to[out=0,in=0,looseness=1.25] (BA);
\draw[arc arrow=to pos 0.55 with length 3mm] (BA) to[out=180,in=-180,looseness=1.25] (AB);
% bottom right
\draw[arc arrow=to pos 0.55 with length 3mm] (BC) to[out=0,in=0,looseness=1.25] (CB);
\draw[arc arrow=to pos 0.55 with length 3mm] (CB) to[out=180,in=-180,looseness=1.25] (BC);
% middle top
\draw[arc arrow=to pos 0.55 with length 3mm] (BD) to[out=90,in=90,looseness=1.25] (DB);
\draw[arc arrow=to pos 0.55 with length 3mm] (DB) to[out=-90,in=-90,looseness=1.25] (BD);
% straight arrows
\draw[postaction={decorate},decoration={markings,
mark=at position 0.45 with {\arrow{Stealth[length=3mm]}}}] (AB) -- (BC);
\draw[postaction={decorate},decoration={markings,
mark=at position 0.45 with {\arrow{Stealth[length=3mm]}}}] (CB) -- (BA);
% upper arcs
\draw[arc arrow=to pos 0.55 with length 3mm] (CB) to[out=180,in=-90,looseness=1.25] (BD);
\draw[arc arrow=to pos 0.55 with length 3mm] (DB) to[out=270,in=0,looseness=1.25] (BA);
% lower right arc
\draw[arc arrow=to pos 0.35 with length 3mm] (DB)
to[out=-90,in=120,looseness=1.25] ($(B.east)+(3mm,0)$)
to[out=-60,in=180,looseness=1.25] (BC);
\draw[arc arrow=to pos 0.85 with length 3mm] (AB)
to[out=0,in=-120,looseness=1.25] ($(B.west)+(-3mm,0)$)
to[out=60,in=-90,looseness=1.25] (BD);
\end{scope}
\end{tikzpicture}
\end{document}
I kept the original solutions for comparison. And yes, one could make the arcs in the above picture really arcs of a circle or an ellipse. You only need the path constructions from what follows.
\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{calc} % shapes,arrows,shapes.multipart,positioning, fit, were not used
\usetikzlibrary{arrows.meta,bending} %< added
\begin{document}
\begin{tikzpicture}
%roads: grey graph
\begin{scope}[every node/.style={circle,thick,draw}]
\node (A) at (0,0) {A};
\node (B) at (3,0) {B};
\node (C) at (6,0) {C};
\node (D) at (3,3) {D};
\end{scope}
\begin{scope}[every edge/.style={draw=lightgray,line width=3pt}]
\path [-] (A) edge (B);
\path [-] (B) edge (C);
\path [-] (B) edge (D);
\end{scope}
%line graph
\begin{scope}[every node/.style={circle,thick,draw,scale=.7}]
\node (AB) at (1.5,-.5) {};
\node (BA) at (1.5,.5) {};
\node (BC) at (4.5,-.5) {};
\node (CB) at (4.5,.5) {};
\node (BD) at (3.5,1.5) {};
\node (DB) at (2.5,1.5) {};
\end{scope}
\begin{scope}[line width=1pt]
% bottom left
\draw let \p1=($(AB)-(BA)$) in (AB) arc(-90:90:{veclen(\x1,\y1)/2});
\draw[-{Stealth[length=3mm,bend]}] let \p1=($(AB)-(BA)$) in (AB) arc(-90:0:{veclen(\x1,\y1)/2});
\draw let \p1=($(BA)-(AB)$) in (BA) arc(90:270:{veclen(\x1,\y1)/2});
\draw[-{Stealth[length=3mm,bend]}] let \p1=($(AB)-(BA)$) in (BA) arc(90:180:{veclen(\x1,\y1)/2});
% bottom right
\draw let \p1=($(BC)-(CB)$) in (BC) arc(-90:90:{veclen(\x1,\y1)/2});
\draw[-{Stealth[length=3mm,bend]}] let \p1=($(BC)-(CB)$) in (BC) arc(-90:0:{veclen(\x1,\y1)/2});
\draw let \p1=($(CB)-(BC)$) in (CB) arc(90:270:{veclen(\x1,\y1)/2});
\draw[-{Stealth[length=3mm,bend]}] let \p1=($(BC)-(CB)$) in (CB) arc(90:180:{veclen(\x1,\y1)/2});
% middle top
\draw let \p1=($(BD)-(DB)$) in (BD) arc(00:180:{veclen(\x1,\y1)/2});
\draw[-{Stealth[length=3mm,bend]}] let \p1=($(BD)-(DB)$) in (BD) arc(00:90:{veclen(\x1,\y1)/2});
\draw let \p1=($(DB)-(BD)$) in (DB) arc(-180:0:{veclen(\x1,\y1)/2});
\draw[-{Stealth[length=3mm,bend]}] let \p1=($(BD)-(DB)$) in (DB) arc(180:270:{veclen(\x1,\y1)/2});
% straight arrows
\draw[-{Stealth[length=3mm]}] let \p1=($(AB)-(BC)$) in (AB) --
++({veclen(\x1,\y1)/2},0);
\draw (AB) -- (BC);
\draw[-{Stealth[length=3mm]}] let \p1=($(CB)-(BA)$) in (CB) --
++(-{veclen(\x1,\y1)/2},0);
\draw (CB) -- (BA);
% upper right arc
\draw[-{Stealth[length=3mm]}] let \p1=($(CB)-(BD)$) in (CB)
arc(270:210:{veclen(\x1,\y1)/sqrt(2)});
\draw let \p1=($(CB)-(BD)$) in (CB)
arc(270:180:{veclen(\x1,\y1)/sqrt(2)});
% upper left arc
\draw[-{Stealth[length=3mm]}] let \p1=($(BA)-(DB)$) in (DB)
arc(00:-60:{veclen(\x1,\y1)/sqrt(2)});
\draw let \p1=($(BA)-(DB)$) in (DB)
arc(00:-90:{veclen(\x1,\y1)/sqrt(2)});
% lower right arc
\draw[-{Stealth[length=3mm]}] let \p1=($(BC)-(BD)$) in (BC)
arc(270:210:{abs(\x1) and abs(\y1)});
\draw let \p1=($(BC)-(BD)$) in (BC)
arc(270:180:{abs(\x1) and abs(\y1)});
% lower left arc
\draw[-{Stealth[length=3mm]}] let \p1=($(AB)-(DB)$) in (DB)
arc(00:-45:{abs(\x1) and abs(\y1)});
\draw let \p1=($(AB)-(DB)$) in (DB)
arc(00:-90:{abs(\x1) and abs(\y1)});
\end{scope}
% unfortunately we need to refill the circles
% alternatively we could have worked with west, east etc. but that's more
% effort and also would distort the circles of the arrows
\foreach \X in {AB,BA,BC,CB,BD,DB}
{\fill[white] (\X) circle (1mm);}
\end{tikzpicture}
\end{document}
The code looks lengthy (OK, it is lengthy ;-) but it is just a copy & paste of one building block. One could write a macro for that...
EDIT: Corrected the direction of the left arcs.
COMMENT: Yes, I know about the decoration.markings library and its ability to put arrows (or whatever) where I want on the path. The reason I am was not using that here is that I don't didn't know how to bend these arrows according to the underlying path, so I just had to draw two paths.
UPDATE: I just realized that two of the arcs were not as you want them. I used the opportunity to write a semi-intelligent macro that does the circular and elliptical arcs. (It is semi-intelligent, so of course you will be able to fool it, but at least in these examples it does its job.) I can't use it for the previously incorrect arcs since they would cross the B node, which looks ugly IMHO.
\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{calc} % shapes,arrows,shapes.multipart,positioning, fit, were not used
\usetikzlibrary{arrows.meta,bending,decorations.markings,intersections} %< added
\newcommand{\DrawArcWithBentArrow}[6][]{%
% #1 option(s)
% #2 start, #3 end, #4 start angle, #5 end angle, #6 angle of arrow position
\pgfmathtruncatemacro{\AngSum}{mod(#4+#5,180)}
\ifnum\AngSum=0
\draw[#1] let \p1=($(#2)-(#3)$) in (#2) arc(#4:#5:{veclen(\x1,\y1)/2});
\draw[-{Stealth[length=3mm,bend]},#1] let \p1=($(#2)-(#3)$) in (#2)
arc(#4:#6:{veclen(\x1,\y1)/2});
\else
\begin{pgfinterruptboundingbox} % make sure the auxiliary paths don't mess up bbox
\path let \p1=($(#2)-(#3)$), \n1 = {veclen(\x1,\y1)/2} in
\pgfextra{\xdef\ArrDist{\n1}};
\path[name path=aux1] ($(#2)+(#4:{-2*\ArrDist})$) -- ($(#2)+(#4:{2*\ArrDist})$);
\path[red,name path=aux2] ($(#3)+(#5:{-2*\ArrDist})$) -- ($(#3)+(#5:{2*\ArrDist})$);
\path[name intersections={of=aux1 and aux2, by=aux3},red]
let \p1 = ($(aux3)-(#2)$), \p2 = ($(aux3)-(#3)$), \n1 = {veclen(\x1,\y1)},
\n2 = {veclen(\x2,\y2)} in
\pgfextra{\xdef\ArrDistA{\n1}\xdef\ArrDistB{\n2}}
arc(#4:#5:{(abs(veclen(\x1,\y1))} and {abs(veclen(\x2,\y2))});
\end{pgfinterruptboundingbox}
\draw[#1] (#2) arc(#4:#5:{\ArrDistA} and {\ArrDistB});
\draw[-{Stealth[length=3mm,bend]},#1] (#2) arc(#4:#6:{\ArrDistA} and {\ArrDistB});
\fi
}
\begin{document}
\begin{tikzpicture}
%roads: grey graph
\begin{scope}[every node/.style={circle,thick,draw}]
\node (A) at (0,0) {A};
\node (B) at (3,0) {B};
\node (C) at (6,0) {C};
\node (D) at (3,3) {D};
\end{scope}
\begin{scope}[every edge/.style={draw=lightgray,line width=3pt}]
\path [-] (A) edge (B);
\path [-] (B) edge (C);
\path [-] (B) edge (D);
\end{scope}
%line graph
\begin{scope}[every node/.style={circle,thick,draw,scale=.7}]
\node (AB) at (1.5,-.5) {};
\node (BA) at (1.5,.5) {};
\node (BC) at (4.5,-.5) {};
\node (CB) at (4.5,.5) {};
\node (BD) at (3.5,1.5) {};
\node (DB) at (2.5,1.5) {};
\end{scope}
\begin{scope}[line width=1pt]
% bottom left
\DrawArcWithBentArrow{AB}{BA}{-90}{90}{0}
\DrawArcWithBentArrow{BA}{AB}{90}{270}{180}
% bottom right
\DrawArcWithBentArrow{BC}{CB}{-90}{90}{0}
\DrawArcWithBentArrow{CB}{BC}{90}{270}{180}
% middle top
\DrawArcWithBentArrow{BD}{DB}{0}{180}{90}
\DrawArcWithBentArrow{DB}{BD}{180}{360}{270}
% straight arrows
\draw[postaction={decorate},decoration={markings,
mark=at position 0.45 with {\arrow{Stealth[length=3mm]}}}] (AB) -- (BC);
\draw[postaction={decorate},decoration={markings,
mark=at position 0.45 with {\arrow{Stealth[length=3mm]}}}] (CB) -- (BA);
% upper right arc
\DrawArcWithBentArrow{CB}{BD}{270}{180}{210}
% upper left arc
\DrawArcWithBentArrow{DB}{BA}{0}{-90}{-60}
% lower right arc
\draw (DB) to[out=-90,in=120] ($(B.east)+(3mm,0)$)
to[out=-60,in=180] (BC);
\draw[-{Stealth[length=3mm]}] (DB) to[out=-90,in=120] ($(B.east)+(3mm,0)$);
% lower left arc
\draw (AB) to[out=00,in=-120] ($(B.west)+(-3mm,0)$)
to[out=60,in=-90] (BD);
\draw[-{Stealth[length=3mm]}] (AB) to[out=00,in=-120] ($(B.west)+(-3mm,0)$);
\end{scope}
% unfortunately we need to refill the circles
% alternatively we could have worked with west, east etc. but that's more
% effort and also would distort the circular shape of the arrows
\foreach \X in {AB,BA,BC,CB,BD,DB}
{\fill[white] (\X) circle (1mm);}
\end{tikzpicture}
\end{document}
The code to produce this result:
