Why is Octahemioctahedron topologically a torus? I'm afraid, that I have a very bad space vision, because I don't see, that  Octahemioctahedron is topologically a torus. Could somebody explain it for me, why is it?


Scene 2. 
@aes:
Finally, with your help I managed to match the vertices of the polyhedron with the vertices on the topological net, thank you. The result is this:


 A: As drawn here it isn't topologically a torus. It isn't even a manifold. But if you think of it as the image of a map from a polygonal complex built out of eight triangles and four hexagons, with vertices and edges corresponding to those shown in your picture (i.e. no vertex at the center, and no edges where the hexagons in your picture meet) then that polygonal complex is a torus.
This is difficult to visualize, but wikipedia has an image which is a sliced open torus (if you identify the opposite edges, you get a torus):

If you compare carefully the way the polygons fit together in the octahemioctahedron you display with this picture (remembering that the opposite edges in this picture are supposed to be identified) you can see they match up exactly. That is, the polygons in each and the way their edges & vertices are identified match up.
This shows the polygonal complex has the structure of a manifold (you can also check this without checking it's the torus just by making sure each edge has two polygons meeting at it and that if you trace around each vertex you get a single loop of polygons) and that the manifold is a torus.
The Euler characteristic method (12 vertices - 24 edges + 12 faces = 0) mentioned in the other answer is also a good one, but you should make sure to check it's a manifold (as in the parenthetical above) and that it's orientable. Then you can conclude without actually visualizing the torus that it is in fact one, as the only oriented closed 2-manifold with zero Euler characteristic is a torus.
A: Foolproof method: use the triangulation shown in your picture to show that the Euler characteristic is 0.
