To understand how a torus can be flat, it's best to replace the idea of folding with the idea of placing portals on edges. Start with a square and put portals between the north and south edges and between the left and right edges. Intuitively this is flat, and this intuition does indeed capture the mathematical notion that a torus is flat.
Take two circles, side by side, connected by an infinitesimally small overlap, and put portals all around the circles to their equivalent point in the other circle. Then fold this over and inflate it into a sphere.
If you're suspicious about the discontinuity where they overlap, I think that's a red herring - no one is claiming the sphere is isomorphic to two disjoint surfaces of any shape.
Alternatively, make it a cube. I can definitely fold a single piece of paper into a cube, or equivalently, I can put portals on my piece of paper to give it the topology of a cube. Intuitively that is flat. But the cube isn't one of the 18 forms proposed for the universe. Nor is any of the other millions of 3d shapes I can make with the same process.
In the two-discs sphere construction, a circle centered on the disc boundaries will appear half in each disc. But it will be short of the usual circumference due to the shape of the discs. All the curvature has been pushed to these boundaries.
In the cube, a circle centered at a vertex will appear on three sides; its circumference will be three-quarters of the usual value. Note that all the curvature of the cube is at the vertices, not the edges!
For the portal-torus, try drawing a circle anywhere, passing through the portals... it will have the usual circumference, zero curvature everywhere.
For the cube, at each vertex you've lost a quarter circle, and there are 8 vertices -- hence the Euler characteristic of a cube is 2.
For the two-disk model of the sphere, a similar thing should be true, I think, but I haven't worked it out in detail -- the integral of "circles lost" over the sphere (the support of this integral is the shared boundary of the disks) should be 2 as well.
This is the Gauss-Bonnet theorem.
For the two-disc sphere, I can't think of an intuitive way to "see" the "circles lost" integral. But here's a different intuitive way to see the total curvature.
Another way to measure the curvature is to look at how much the sum of the interior angles of an n-sided polygon exceeds the usual sum π(n - 2). It's most common to think about triangles, but we can also think about 2-gons... these are usually degenerate shapes with a sum of interior angles of zero.
But on the two disc-sphere, draw two lines, each from the center of one disc to the center of the other disc, passing straight through the glued boundary. These form a 2-gon with sum of interior angles (and also excess over the usual value) equal to twice the angle between the lines. To get the total curvature of the whole sphere, let each of the two interior angles be 2π, for a total of 4π... two circles, the Euler characteristic.
One thing that helped me is considering how to deform the edges of your glued circle examples into a flat square. First, map the two sides of the seam two the two corners of the plane. This becomes a sphere by folding it into a triangle and then gluing (or portaling) up neighboring edges. Having the portals on neighboring edges means that they change directions in weird ways.
This is somewhat different than the torus with portals example, where the portals were placed on opposite edges of the plane.
Edit: Found some diagrams!
https://en.wikipedia.org/wiki/Fundamental_polygon#Examples_o...
Parallel transport is broken in your model of the sphere. Take this example. Take a vector pointing north in circle 1, and send through the north portal. It should be pointing south when it gets to the other circle. Fine -- north in circle 1 corresponds to south in circle 2.
Now send the north-pointing vector east instead. It is going to point north in circle 2.
So the north vector changes direction depending on which portal it goes through. So parallel transport changes directions. Hence your model of the sphere is not flat.
I wouldn't say the curvature implied by this property is "intuitively" 0 - why not 1, or 360°, or 180°? - but I can see how it's an interesting property that only a few shapes can have.