Orientability of projective space

According to Poincare Duality, a compact $n$-manifold $M$ is orientable iff $H_n(M,\mathbb{Z}) \neq 0$. The homology groups of real projective space can be computed using a cell decomposition with only one cell in each dimension. This wikipedia article has a nice explanation of why this leads to a vanishing top dimensional homology group or $\mathbb{R} \mathbb{P}^n$ iff $n$ is even.

An orthonomal frame at the north pole of the 2 sphere can be brought to the south pole either by parallel translation along a great circle or by the antipodal map. If the projective plane were orientable then these two resulting frames would be the same.

But the translated frame and the identified frame have opposite orientation.

For odd spheres they have the same orientation so the odd projective spaces are orientable.

As you mentioned in the question, $\pi:\mathbb{S}^n\rightarrow\mathbb{R}\mathbb{P}^n$ is a $\mathbb{Z}_2$-principal covering, with deck group $\{1,\alpha\}$, where $\alpha$ is the antipode map of $\mathbb{S}^n$. Since $\pi\circ\alpha=\pi$ we get pullbacks satisfying $\alpha^*\circ\pi^*=\pi^*$, which shows if $\mathbb{R}\mathbb{P}^n$ is orientable then $\alpha$ must be orientation preserving by pullbacking the same orientation on $\mathbb{R}\mathbb{P}^n$ to $\mathbb{S}^n$. But the antipode map $\alpha$ is orientation preserving iff $n$ is odd.