If $f \circ \sin$ is measurable, then we need to show that $(f \circ \sin)^{-1}(B) = \sin^{-1}(f^{-1}(B))$ is measurable for every Borel set $B$, it is sufficient to show that $\sin^{-1}$ takes a Lebesgue set into a Lebesgue set. We have also tried to work with the fact that $\arcsin$ is locally Lipschitz, so if this property is sufficient to show that the image of a Lebesgue set is Lebesgue, then $\sin^{-1}(f^{-1}(B)) = \cup_{n \in \mathbb{Z}} \arcsin(f^{-1}(B)) + 2n\pi $ would measurable.
It could happen that $f\circ \sin$ is not measurable, but the task of finding a set $B$ such that $\sin^{-1}(f^{-1}(B))$ is not measurable seems very difficult (maybe impossible?).
Yes, $ f \circ \sin$ is measurable. The composition of measurable functions is in general not measurable, so there is something to do. You allude to using Lipschitzness, but I think the relevant property is absolute continuity (which sine satisfies). We just need to show that absolutely continuous functions take Lebesgue measurable sets to Lebesgue measurable sets. This is answered here: Absolutely continuous maps measurable sets to measurable sets. The basic point is that absolutely continuous functions take null sets to null sets.