The theorem says that the G-side-pairing used for "glueing" such a manifold should be proper. Proper means that each cycle under the side-pairing is finite and has a solid angle sum of $4 \pi$. Now, the question is to determine what kind of cycles and solid angles you've got in the picture.
If you consider a point inside this tetrahedron, then there is no problem to circumscribe a small sphere around it and thus the solid angle will be $4 \pi$. The respective cycle consists of one point.
If you take a point on a face (inside that face), then you automatically have a half-sphere with the centre at this very point (the solid angle is $2 \pi$). However, this point is identified with some other point on the second face. The point there also has a half-sphere. Thus, the respective cycle consists of two points. And the solid angle is $2\pi + 2\pi = 4 \pi$ again.
Now the most delicate thing is to verify what happens to a point sitting at an edge (inside it) or a point that coincides with one of the vertices. Actually, you've got to know the respective dihedral angles to check if the solid angle in question is $4\pi$. Even if you know the dihedral angles, this task needs some labour.
However, by Poincare's polyhedron theorem, you have to verify that the dihedral angle sum over each equivalence class of edges is $2\pi$. Check Ratcliffe's book for the theorem and its proof. The theorem is pretty nice and often in use. Also there are examples of how to glue the figure-eight complement out of regular ideal hyperbolic tetrahedra (in the same book, sorry for the lack of exact references to paragraphs). This could be a good illustration on how to check the properness of a side-pairing (may be even without Poincare's theorem).
Edit: For example, here you say that AC is always glued to itself and there are no other edges in its equivalence class under the glueing (am I right here?). Thus, there is only one equivalence class in the cycle for every point sitting on that edge. Thus, the respective angle sum from Poincare's theorem equals the dihedral angle along AC. Supposedly, this angle is never $2\pi$, since you may want your tetrahedron to be convex (although you do not specify the angles in your question). Thus, you will never have a manifold. Same answer "not a manifold" if you try to compute the solid angle for a point from the interior of the edge AC.
Hope my reply help a bit. Cheers!