Your non-vanishing sphere test is equivalent to calculating the homotopy group of the embedding space, with the shape taken out and an extra point at infinity, so the sphere can vanish either by shrinking to a point or by expanding to infinity. Since R^n with a point at infinity is homeomorphic to an n-sphere, you're really talking about the holes of the complement of a shape embedded on a sphere.
A surface that crosses itself is not a sphere, since you can use the crossing to create simpler loops that do the same function
Well you can define a sphere like that if you want, but you'll find it's not as useful as the standard definition. A sphere in topology is simply the image of some continuous map from the geometric sphere onto the space. It can cross itself in the space, because in the whole space (preimage x image) it doesn't cross because the parameters are different. You'll find it very difficult to talk about holes of the projective plane and klein bottle.
The non-vanishing sphere test is a reworked version of homotopy and patches are equivalent to homology if you do it right, but homotopy and homology are irreconcilably different. On the other hand, the non-vanishing sphere test is done in the embedding space instead of in the shape, so maybe it won't have these problems
. So possibly the non-vanishing sphere test is equivalent to the method of patches. I wouldn't bet on it though. I really like the missing patch theory, but I don't think we need to mess around with nonstandard homotopy theories.
In terms of the other two patches, the patch must be mounted: that is, its surface must be made of elements that exist on the figure it is to be mounted on. Since the intersection between the elements are the surfaces of the holes, the choric hole needs a hedric surface to become attached, and the hedric hole needs a latric surface. These intersect in a point. In this way, all of the possible holes in the surface of the torus are closed too.
Ok, I get the idea of mounting a patch onto its boundary. But you have still pockets left behind. Let's go back to a standard torus. You fill in the middle hole with a disk mounted on a circle. Then you put another disk inside the tube. But there's still space inside. You have to fill it with a ball. You can't completely fill in the torus with two disks. Nor can you forget the second disk, because a ball won't fill up the whole tube or else it's no longer a ball. So the torus requires two disk patches and a ball patch. Same goes for the duocylinder 22 since they're homeomorphic. Do we agree on this?
Similarly, you can't fill in 32 without one disk, one ball and a gongyl (solid 4D sphere).