For the 4-gon, you have exactly these four cases. The reason is that we do disallow discontinuities, i.e. we are interested in surfaces = twodimensional manyfolds: In the surface, the (identified) sides become a single curve that locally looks like the $x$-axis within $\mathbb R^2$, i.e. there must be exactly two sides (corresponding to lower and upper half plane). This means that exactly two edges must be labelled "A" etc. We might allow the two A edges to be adjacent with "wrong" directions. But such cases be better resolved by splitting the two edges in half (and have a "legal" (n+2)-gon). Indeed, you may note that the projective plane can be considered as coming from an "illegal" 2-gon this way (while the sphere might also be considered to actually be a glued 2-gon) With these restrictions, we have only few combinatorical possibilities: If the "red" edges are adjacent, we necesarily obtain the sphere. If they are not adjacent, we have one of the three remaining situations: Both colour pairs are "parallel", or both are "antiparallel" or we have a mixed case.
We can derive a simple upper bound for the topologies possibly constructed from the 2n$-gon: There are $(2n-1)\cdot(2n-3)\cdot\ldots\cdot3\cdot1=\frac{(2n)!}{2^nn!}$ possibilities to partition the $2n$ edges into pairs (fix the first, select his partner among the rest, fix the first available, select his partner among the rest, etc.). For each such partition, we may label each pair either "parallel" or "antiparallel". Thus we end up with $\frac{(2n)!}{n!}$ as an upper bound. Note that the correct number is a lot smaller (for example we obtain $\frac{4!}{2!}=12$ instead of $4 for the 4-gon):
- If adjacent edges have the same label, only one orientation is allowed
- Rotating the polygon produces the same topology (the rotation induces a homeomorphism)
- Some of the topologies might be obtainable from a 2k$-gon with $k