Longer Answer :
The change of variables formula(COV) does not apply in the given conditions.
Think of the change of variables formula like this : given a set whose area you cannot find, you change it in a manner such that its area does not change and it assumes some possibly nice shape (whose area you can calculate), and the COV formula gives you license to do this, but the manner in which you "change" the set cannot be arbitrary. For example, imagine you are finding the area of a rhombus. You can "rotate and straighten" it without changing the area and get a rectangle whose area you know, and the COV formula will tell you the area of the rhombus is the area of the rectangle. However, if I decide to fold the Rhombus over itself, then I get a shape of smaller area (a triangle), so clearly the COV should not apply here, because the area has changed!
So, I have got to state by COV carefully. The most important thing, as you may have noticed, is this : the change should be "invertible". That is, the map should be injective i.e. one-one and surjective i.e. onto the new shape. The map "rotating and straightening the Rhombus" is invertible because we can just "tilt the sides and rotate back" to obtain the old shape. The map "folding the Rhombus" is not injective, because when you fold you are sending two different points to the same new point (like how when you fold paper, some points come on top of each other), and so you cannot "invert" the operation, because given the triangle the inverse map to the Rhombus maps a point to more than one point, and these are not functions, as you know. So COV would not apply there.
Now, the map $\sin : [0,2\pi] \to [-1,1]$ is not injective, for example $\sin \pi = \sin 0 = 0$. Therefore, the change of variable is not injective!
Which is exactly why you face such absurdity : the formula does not apply here. What Yves said is that the statement $\sin x = t$ implies $x = \arcsin t$ is false. Indeed, this is the case.
Because, what is $\arcsin 0$, for example? Is it $0$, because $\sin 0 = 0$? Is it $\pi$, because $\sin \pi = 0$? It cannot be both, right? Suppose it is one of them, so for example if I say that only $\arcsin 0 = 0$, then the statement $\sin x = t \implies x = \arcsin t$ will be false because $\sin \pi = 0$ but $\arcsin 0 = 0 \neq \pi$!
That is why your example is failing. Graphically speaking, you have "folded" the graph of $f(x)$ over itself (as the many preimages of the sine would describe), and the folded graph will have area zero.
So, why does "breaking the integral" work?
Well, it is because $\sin$ on those blocks is injective! For example, the function $\sin x $ on $[0,\frac \pi 2]$ is injective. Similarly on $[\frac \pi 2 ,\pi]$. So here, the function $\arcsin$ will be well-defined with unique preimage on each of these blocks, so COV will apply.
You have to now think of it this way : imagine you are finding the area of a shape. Maybe this shape is too difficult to work with, so you break it into parts. If each of these parts can be transformed injectively into a nice-enough shape, then COV will apply, you can find the area of those shapes and then add them up. For example, this can be done with a disjoint union of rhombuses (rhombi?), so you don't know how to work with the union, but you look at each rhombus, rotate and tilt the sides to get a rectangle and find the area of that.
So COV may not work on the whole shape, but maybe we can break it into parts on which the COV will work, and then add up the results.
In conclusion, your confusion is with a catch in the COV, and you have to be careful in using it. Always ensure your transformation is injective!