I understand that there are two solutions to $y = \sin x$, one for the case $y>0$ and $y<0$, the solutions being $(0, \arcsin(y)),(\pi-\arcsin(y),π)$ for $y>0$ and $(\pi-\arcsin(y),2\pi+\arcsin(y))$ for $y<0$.
Found that here on page 4.
I don't understand how to combine the solutions of the two cases in order to get the final distribution function of $Y$.
It is helpful to draw a diagram. Here we have a unit circle, and the random variable $Y = \sin (X)$ is the $y$-coordinate of a random point on the unit circle generated by taking an angle $X \sim U(0,2\pi)$ from the origin.
In the case where $y\geq 0,$ we have $Y > y$ if and only if the angle is between $[\arcsin(y), \ \pi - \arcsin(y)],$ so $$\mathbb{P}(Y \leq y) = 1 - \frac{ (\pi - \arcsin(y)) - \arcsin(y)}{2\pi} = \frac{ \pi + 2\arcsin(y)}{2\pi}.$$
In the case where $y<0,$ we have $Y\leq y$ if and only if the angle is between $[\pi - \arcsin(y), \ 2\pi + \arcsin(y) ],$ so
$$\mathbb{P}(Y\leq y) = \frac{ (2\pi + \arcsin(y)) - ( \pi - \arcsin(y))}{2\pi}$$ $$ = \frac{\pi + 2\arcsin(y)}{2\pi}$$
So we get the same expression for $\mathbb{P}(Y\leq y)$ for both cases of $y\geq 0, y<0.$
For $y>0$ you can write$$\Pr\{Y<y\}{=1-\Pr\{Y>y\}\\=1-\Pr\{\sin X>y\}\\=1-\Pr\{X\in( \sin^{-1}y,\pi-\sin^{-1}y)\}\\=1-{\pi-2\sin^{-1}y\over 2\pi}\\={\pi+2\sin^{-1}y\over 2\pi}}$$and for $y<0$ similarly $$\Pr\{Y<y\}={\pi+2\sin^{-1}y\over 2\pi}$$so that$$\Pr\{Y<y\}={\pi+2\sin^{-1}y\over 2\pi}\quad,\quad y\in[-1,1]$$and $$f_Y(y)={1\over \pi\sqrt{1-y^2}}\quad,\quad -1<y<1$$
