No: Consider $\mathbb{Z_{2}}$ acting on the universal cover $X=S^{n}$ of $Y=\mathbb{RP}^{n}$ for $n$ odd. In this case both $X$ and $Y$ are closed and orientable, so we can talk about Poincare Duality and transfer maps. The map $\pi^{*}$ must be zero when restricted to the certain degrees of the cohomology groups, as $H^{k}(Y;\mathbb{Z})=\mathbb{Z}_{2}$ and $H^{k}(X;\mathbb{Z})=0$ for $k$ even with $0 < k < n$.

Yes: Let $\alpha$ be a cohomology class in $Y$. For simplicity, let's assume that $PD_{Y}\alpha$ is represented by an embedded cycle in $Y$. Then $PD_{X}\pi^{*}\alpha=\pi^{-1}PD_{Y}(\alpha)$. Therefore, $\pi\circ PD_{X}\pi^{*}\alpha = \pi (\pi^{-1}PD_{Y}\alpha)=|G|PD_{Y}\alpha$ as the map $\pi: \pi^{-1} PD_{Y}\alpha\rightarrow PD_{Y}\alpha$ is a $|G|$-to-1 covering. Now apply $PD_{Y}^{-1}$ to the previous equation to obtain $\pi_{!}\circ \pi^{*}\alpha=|G|\alpha$.

No: Consider $\mathbb{Z_{2}}$ acting on the universal cover $X=S^{n}$ of $Y=\mathbb{RP}^{n}$ for $n$ even. As $H^{n}(Y;\mathbb{Z})=\mathbb{Z}_{2}$ and $H^{n}(X;\mathbb{Z})=\mathbb{Z}$, the map $\pi^{*}$ must be zero when restricted to the $n$th cohomology groups.

No: Consider $\mathbb{Z_{2}}$ acting on the universal cover $X=S^{n}$ of $Y=\mathbb{RP}^{n}$ for $n$ odd. In this case both $X$ and $Y$ are closed and orientable, so we can talk about Poincare Duality and transfer maps. The map $\pi^{*}$ must be zero when restricted to the certain degrees of the cohomology groups, as $H^{k}(Y;\mathbb{Z})=\mathbb{Z}_{2}$ and $H^{k}(X;\mathbb{Z})=0$ for $k$ even with $0 < k < n$.