All gluings result in a compact surface. It's obvious that every point in the interior of the polygon is locally homeomorphic to $\mathbb{R}^2$, and it's not so hard to see that this is also true of the interior of the edges, so it only remains to inspect the vertices, and this isn't so bad either: by "following the gluing" clockwise or counterclockwise you eventually produce a cycle that closes up a neighborhood of each vertex. This is easier for me to see geometrically than to spell out rigorously.
I know of two straightforward algorithms to determine the homeomorphism type of the corresponding surface. The first uses Mayer-Vietoris to compute the first homology $H_1(M, \mathbb{Z})$, which it's known completely determines the surface. The second is combinatorial and takes some diagrams to explain properly, but is totally elementary; it should be explained anywhere the classification of (compact) surfaces is proven in the usual way.
I am adding the answer I received by email from Doctor Tadashi Tokieda, he is the director of studies in mathematics at Trinity Hall, University of Cambridge, some bio here and here, I was following his Topology and Geometry open lectures at Youtube for the AIMS, so I dared to send him an email yesterday (I did not expect an answer, just tried) and received his answer!! (thank you very much Doctor Tokieda, it was a honor!)
He sent me a very nice easy to follow explanation of the concepts, so I am transcribing exclusively the explanation here as a complement (more adapted to layman terms) to the good answer of @AloizioMacedo. The drawings are mine, so I apologize if they are not very accurate. Here it is:
Let us take the example of a disk. Think of the disk as a billiard table and of its boundary circle as the cushion rails. If I understood aright, your picture is: shoot a particle, it slides along a straight line, but when it hits a cushion rail, it reflects, and slides along another straight line... Capture this picture by some sort of gluing diagram.
How about this? Take two copies of the disk, $A$ and $B$. Sew $A$ and $B$ together along their boundary circles, like a pita bread. Puff inside a little, so that the resulting closed surface $S$ looks like a very squashed sphere, with $A$ as the northern hemi-pita and $B$ as the southern hemi-pita. At every point along the equator of $S$, the surface is smooth and its tangent plane is vertical.
Imagine a particle sliding on $S$, and watch it from "above". While it is sliding on $A$, it appears to travel along more or less a straight line on a disk. As the particle approaches the equator, crosses the equator, and gets into $B$, you see its image from above approach the boundary circle of the disk, reflect off the boundary, and continue traveling on the disk more or less along a straight line. We had to say "more or less", because the curved $A$ and $B$ mean that the images of the trajectories seen from above are not exactly straight. But you agree that, as we squash $S$ more or more flat, the trajectories converge to straight lines. In summary, the picture you want on a domain (our example was a disk) can be realized by doubling the domain into a closed surface. The dynamical behavior you want on the domain is the projection of the dynamics on that surface.
There is a whole sub-field of mathematics called "billiards". There is also a well-studied topic of geodesics (trajectories of sliding particles) on Riemannian manifolds. Our discussion above shows that you can reduce the former to the latter, by taking the "squashing" limit.
See: Dynamical billiards at Wikipedia.
Best Answer
It's the real projective plane. Its Wikipedia page provides diagrams much the same as yours.