I get 12 also. The easiest way to see it seems to be to colour the original edges of the polygon in red, and the inner lines in blue.
This makes n red lines and n-3 blue lines, separating the polygon into n-2 triangles. At least two triangles will have two red edges (one on either side of any blue line). This leaves n-4 triangles and n-4 red lines, so each of the other triangles each have exactly one red line.

Once this is figured out, it is easy to derive a formula, by starting from a red-red-blue triangle and building up to the polygon.
For instance with 6 vertices, the procedure is as follows:

• Start with a red-red-blue triangle. There are 6 ways to label the vertex incident to the two red lines. We will label the other vertices clockwise when the polygon is completed.
  
  
• Put a new triangle on the other side of the blue line. You must colour one of its edges red and the other one blue. There are two ways to do this.
  
  
• Again put a new triangle on the other side of the blue line. You must colour one of its edges red and the other one blue. There are again two ways to do this.
  
  
• You are now up to 5 vertices. Put a new triangle on the other side of the blue line, colour both of its edges red, and you are done.

All that remains is to label the vertices clockwise around the original labeled vertex, and perhaps stretch the polygon if you want it to be regular.

There are $6 \times 2 \times 2 = 24$ ways to build a triangled polygon in this way. But each polygon has been counted twice, because there were two choices for the original red-red-blue triangle. So there are indeed 24/2 = 12 ways do divide the polygon, as you found out. This can be generalised to any number of sides.

Claude