In the second example I am trying to work the the library intersection. It works fine as long as I don't use the calc library.
The first example shows my intention. I am using ($2*(w1')$) which produces the expected result:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections,calc}
\begin{document}
\begin{tikzpicture}[x=3cm,y=3cm]
\coordinate (O) at (0,0);
\coordinate (F') at (-0.8,-0.1);
\coordinate (F'') at (0.8,0);
\draw[thick](O) ++(F') ++ (F'');
\draw[gray,name path=circle around F'] (F') circle (1.2);
\draw[gray,name path=circle around F''] (F'') circle (1.2);
\draw[name intersections={of=circle around F' and circle around F'',by={w1,w1'}}] (intersection-1) node[dot,draw]{}
-- (intersection-2)node[dot,draw]{};
\draw[red,thick] (w1) -- (w1');
\draw[blue,thick] (w1) -- ($2*(w1')$);
\end{tikzpicture}
\end{document}
In the picture you can see that the blue line is expanded.

In the second (not minimal) example it fails:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections,calc}
\begin{document}
\tikzset{dot/.style={circle,minimum size=3pt,inner sep=0pt,outer sep=0pt,thick,fill,}}
\begin{tikzpicture}[x=3cm,y=3cm]
\coordinate (B) at (-1,0);\node[dot] at (B) {};\node[left] at (B) {$B$};
\coordinate (D) at (0,1);\node[dot] at (D) {};\node[anchor=south east] at (D) {$D$};
\coordinate (F) at (0,-.25);\node[dot] at (F) {};\node[anchor=west] at (F) {$F$};
\draw[thick,name path=line FB] (F) -- (B);
\draw[thick,name path=line FD] (F) -- (D);
\draw[gray,name path=circle around F] (F) circle (0.8);
\path[name intersections={of=circle around F and line FB,by=F'}];
\path[name intersections={of=circle around F and line FD,by=F''}];
\draw[gray,name path=circle around F'] (F') circle (0.6);
\draw[gray,name path=circle around F''] (F'') circle (0.6);
\draw[name intersections={of=circle around F' and circle around F'',by={w1,w1'}}]
(w1) node[dot]{}
-- (w1')node[dot]{};
\draw[red] (w1') -- ($2*(w1)$);
\end{tikzpicture}
\end{document}
The red line misses the intersection. I don't know why.

What is the problem?

\draw[red] (w1') -- ($(w1')!2!(w1)$);. So probablycalclibrary is assuming something different for the omitted coordinate in($coord1!2!coord2$)type of input. – percusse Mar 2 '12 at 16:50calclibrary is the deeper magic from beyond the dawn of time. – Andrew Stacey Mar 2 '12 at 16:57