It is a covering if $\pi$ is continuous and onto. Take $y' \in Y'$, we want to find a neighborhood $U_{y'}$ of $y'$ such that the preimage under $\pi$ is the union of some disjoint open sets homeomorphic to $U_{y'}$ under $\pi$.
Let $p_2(y') = x$. There is a neighborhood $V_x$ of $x$, such that
1. $p_2^{-1}(V_x)$ is a disjoint union of open sets, each homeomorphic to $V$ under $p_2$. Call the open set containing $y'$ $U_{y'}$, and $q_2$ to be the inverse (homeomorphism) of $p_2 |_{U_{y'}} : U_{y'} \to V_x$.
2. $p_1^{-1}(V_x)$ is a disjoint union of open sets $W_a$, each homeomorphic to $V$ under $p_1$, and $p_1(a) = x$.
We need $\pi$ to be onto to make sure the preimage $\pi^{-1} (U_{y'})$ isn't empty. If $\pi$ is continuous, then $\pi^{-1}(U_{y'})$ can be checked to be a disjoint union of open sets $W_a$, with $\pi (a) = y'$. Continuity is used to make sure that $W_a$ are mapped to $U_{y'}$, for otherwise it may be possible to split $W_a$ into pieces, each mapping to different covering neighborhood, as shown in the other answer.