Build the new graph as follows. First take two identical copies of the vertices and blue edges of the original graph. (Perhaps it will help to draw an example). The first copy is $G_o$ and the second is $G_e$. Notice that you have no way of moving between the two copies yet, and you can move inside each copy only using the blue edges.
Now add the red edges between the two copies. There is no red edge within a copy - only between them. So now if you start from $G_o$, if you go through an even number of red edges, you are necessarily back in $G_o$. If you want to get from any vertex in $G_o$ to any vertex in $G_e$, you MUST go through an odd number of red edges. So to get from $s_o$ to $t_e$ you must pass through an odd number of red edges. (This is just as true if you want to get from $s_o$ to $s_e$ - they are now completely different vertices in the new graph, and have no relation to each other except in our minds, because we perceive them as two copies of $s$.)
I hope this clarifies everything up.
gal (guest) 26 Dec 2014 11:17