Partial answer

I assume that $[A:C] = [B:D] = 2$, since this may not follow from the isomorphisms $C \simeq A/\{-1,1\}$, $D \simeq B/\{-1,1\}$. 

The morphism $\pi$ is surjective, like the canonical projection on $B/D$, denoted by $p_{B/D}$ in what follows, so $\mathrm{Im}(p_{B/D} \circ \pi) = B/D$. But $\mathrm{Ker}(p_{B/D} \circ \pi) = \pi^{-1}(D)$. Hence, 
$A/\pi^{-1}(D) \simeq B/D$, so $[A:\pi^{-1}(D)] = [B:D] = 2$.