Platonic Solids - Why Five?

One answer: Things don't inevitably match up on the bottom. The tetrahedron is the counterexample. Constructing it based around one vertex as you describe (call it the north pole) does not result in a matching cap around the south pole. Of course, the gap around the south pole is still congruent to the three original triangles. It must be by the rules of Platonic solids, because the other three vertices must each behave identically to the original north pole, meaning the south-pole-centered triangle must be identical to the other triangles.

Here's a more general answer. The solid always has N-way radial symmetry when viewed from above any vertex, such as our north pole. All longitude lines at 360°/N intervals are identical from this method of construction. There is nothing to differentiate the triangle of your icosahedron on the 0° meridian from its counterpart at 72°, so they will all fall into the same relationship with the south pole. The polygons can't overlap the south pole from one direction while falling short from another. In the tetrahedron, the south pole gets overlapped equally from all three radial directions. In the other four solids, the edges and faces meet exactly at the south pole.

I can't answer exactly why that last sentence is true. It reduces to, why does this construction method always produce symmetry across the equator for the four non-tetrahedron solids? Your strip of 10 triangles on the icosahedron: if that is indeed equatorial, then we have both latitudinal and longitudinal symmetry which renders your north and south hemispheres identical so things will match up. But why does that strip of 10 triangles end up centered on the equator? I do not have the answer to that, but maybe someone does.

This seems backwards to me. It's not as though things "line up" magically, but rather there's enough "wiggle room" to allow things line up. It's similar to me to saying that given three non-negative integers A, B, and C, you can form a triangle with side lengths A, B, and C if and only if the sum of any two numbers is greater than the third.

"Why" this is the case is best understood by seeing what happens when it doesn't hold. Or, another way would be to take two line segments attached at the end, fix one, and rotate the other. As you rotate, draw a line between the endpoint of one segment and the other. The range of lengths of this third side is going to be the permissible range of lengths for any third side length and it's going to be "largest" when the two original line segments have a 180º angle between them (i.e., their sum).

But I think for plenty of people this wouldn't suffice to answer the question "why." It could still seem like a "happy accident." This is always the tricky thing with why questions -- what constitutes an explanation for one person is entirely unsatisfactory to another.

Imagine laying out 4 squares the same way. The cap wouldn't be 3D at all, you'd just end up with a big flat square at the top! Now, imagine trying to lay out 5 equilateral triangles in 2D instead of 3D. You'd have a gap. To make the "cap" in 3D, you just pull the first and last triangle together to close the gap, creating a cone. The smaller the gap is in 2D, the less you have to "fudge" the fit in 3D, and the flatter the cap is.

I wouldn't really call the 24-cell "a strange version" of the octahedron. It really isn't anything like any of the 3d Platonic solids. (It does have octahedral faces, though.)

By that method, there's eight. Also included would be the constructions of regular triangles meeting six at a vertex, and regular squares meeting four at a vertex.

That does not work, the sum of infinite zeroes is still zero.

It doesn't have to be infinitely large. It could simply be on a bounded manifold. And as another commenter noted, there are eight by this criterion.

The sixth is a teapot.