Map on homology induced by covering map $S^n \to \mathbb{R}P^n$

Let $R\sigma_n^{(1)}$ denote inclusion of the cell $e_n^{(1)}$ followed by multiplication by $-1$ on $S^n$. Then $p_*(R\sigma_n^{(1)})=p_*(\sigma_n^{(1)})$, so the only thing we have to decide is which of $\sigma_n^{(2)}=\pm R\sigma_n^{(1)}$is true.

As $[\sigma_n^{(1)}+\sigma_n^{(2)}]$ is a homology class, we have $\partial(\sigma_n^{(1)})=-\partial(\sigma_n^{(2)})$.

So if $R\partial(\sigma_n^{(1)})=\partial(\sigma_n^{(1)})$, then $\partial(R\sigma_n^{(1)})=-\partial(\sigma_n^{(2)})$. Then $\sigma_n^{(2)}=-R\sigma_n^{(1)}$ and $p_*(\sigma_n^{(2)})=-p_*(\sigma_n^{(1)})$. Then $$p_*([\sigma_n^{(1)}+\sigma_n^{(2)}])=0.$$

On the other hand, if $R\partial(\sigma_n^{(1)})=-\partial(\sigma_n^{(1)})$, then $\partial(R\sigma_n^{(1)})=\partial(\sigma_n^{(2)})$. Then $\sigma_n^{(2)}=R\sigma_n^{(1)}$ and $p_*(\sigma_n^{(2)})=p_*(\sigma_n^{(1)})$. Then $$p_*([\sigma_n^{(1)}+\sigma_n^{(2)}])=2\tau_n.$$

Multiplication by $-1$ preserves orientation of the equator $S^{n-1}$ precisely when $n$ is even. That is multiplication by $-1$ sends a generator of $H_{n-1}(S^{n-1})$ to $(-1)^{n}$ times itself.

To see this note that if we multiply just one pair of co-ordinates by $-1$, this is a $180^\circ$ rotation, hence homotopic to the identity. Thus when $n$ is even, we have a sequence of homotopies from multiplication by $-1$ to the identity. On the other hand if $n$ is odd, then we have a sequence of homotopies from multiplication by $-1$ to a reflection.

We conclude that $p_*([\sigma_n^{(1)}+\sigma_n^{(2)}])=0$ if $n$ is even and $p_*([\sigma_n^{(1)}+\sigma_n^{(2)}])=2\tau_n$ if $n$ odd.