Gauss-Bonnet theorem for spheres that almost look like a torus [Corrected due to Jason's answer.]
Imagine a torus and a flat disk fitting in the middle of its "hole" (a doughnut with a membrane in the middle). Cut the torus at its inner equator, duplicate the disk, move the two copies away from each other slightly, widen the cut appropriately and join the two flat disks with the sliced torus (anyway you like). 
You get a surface $M$ homeomorphic to the sphere - thus with integral curvature $\int_S\kappa = 4\pi$ - , but with integral curvature equal to that of the torus $\int_T\kappa = 0$ plus a contribution from the "regions of agglutination" where the two disks and the sliced torus are glued together (the disks by themselves having zero curvature).
Is it simply a consequence of the Gauss-Bonnet theorem that however smoothly or abruptly you glue the two disks and the sliced torus together the integral curvature in the "region of agglutination" has to be $4\pi$?
Or is there a mistake in my description of the surface or in my
understanding of the Gauss-Bonnet theorem?
 A: First, note that the integral over the torus is $0$, not $2\pi$ (since the Euler characteristic of a torus is $0$).  I suppose this just makes the result even more surprising.
There are (at least) two things happening which add up to the net increase of $4\pi$ to the total curvature.
First note that you're cutting out exactly the points of largest negative curvature on the torus:  If you imagine revolving a circle around an axis to make the torus, then the outside semicircle rotates to the points with non-negative curvature while the inside semicircle rotations to points with non-positive curvature.  So, in some sense, you've cut out the largest contributer of negative curvature.
Of course, you may object that you're really cutting out a very tiny portion, so it shouldn't affect the answer much, but here is where the other part comes into play.
In order to connect the top half of the torus to the top disc in a smooth way, you have to bend the torus quite bit.  More specifically, imagine rotation $(x-2)^2 + y^2$ around the $z$-axis to get the torus.  And just focus on the semi-semicircle in the top left portion of this circle.  Near the equator, the tangent line is a very large positive number.  In order to connect up with the disc (horizontal tangent), you must bend the graph so this very large number becomes $0$.  That introduces a ton of second derivative (since you're bending the tangent vectors a lot).
More importantly, if you rotate that portion of the graph around, the part near the equator used to be contributing negative curvature, but is now contributing a lot of positive curvature (because it's bent so much).  That's where you really get the $4\pi$ from.
