Note that $\omega$ is an $n$-form on $S^n$, an $n$-dimensional manifold. Therefore $d\omega = 0$ as it is a $(n + 1)$-form on $S^n$. So $\omega$ is closed. Since pullbacks commute with exterior differentiation, we have that $d(f^\ast \omega) = f^\ast d\omega = 0,$ so $f^\ast \omega$ is closed as well. Now apply what you said about $S^{2n-1}$ having trivial cohomology in degree $n$.
Now to show that $\int_{S^{2n-1}} \alpha \wedge d\alpha$ is independent of the choice of $\omega$ and $\alpha$, let $\omega' \in \Omega^n(S^n)$ be another $n$-form on $S^n$ such that $\int_{S^n} \omega' = 1$ and let $\alpha' \in \Omega^{n-1}(S^{2n-1})$ be such that $d\alpha' = f^\ast \omega'$. We will show that $\int_{S^{2n-1}} \alpha \wedge d\alpha = \int_{S^{2n-1}} \alpha' \wedge d\alpha',$ which implies the integral is independent of the choices.
Since $H^n(S^n; \mathbb{R}) \cong \mathbb{R}$, by deRham's theorem there is some $\tau \in \Omega^{n-1}(S^n)$ such that $\omega' = \omega + d\tau.$ Now $d(\alpha' - \alpha - f^\ast \tau) = f^\ast(\omega' - \omega - d\tau) = 0$ and $H^n(S^{2n-1}; \mathbb{R}) \cong 0$, so again by deRham's theorem there exists some $\eta \in \Omega^{n-2}(S^{2n-1})$ such that $\alpha' = \alpha + f^\ast \tau + d\eta.$ Hence we have that \begin{align*} \alpha' \wedge d\alpha' & = (\alpha + f^\ast \tau + d\eta) \wedge (d\alpha + f^\ast d\tau) \\ & = \alpha \wedge d\alpha + \alpha \wedge d(f^\ast \tau) + f^\ast(\tau \wedge (\omega + d\tau)) + d(\eta \wedge (d\alpha + f^\ast d\tau)) \\ & = \alpha \wedge d\alpha + \alpha \wedge d(f^\ast \tau) + d(\eta \wedge (d\alpha + f^\ast d\tau)), \end{align*} where in going to the third line we used the fact that $f^\ast(\tau \wedge (\omega + d\tau)) = 0$ since $\tau \wedge (\omega + d\tau) = 0$ as it is a $(n + 1)$ form on $S^n$. Now, since \begin{align*} \alpha \wedge d(f^\ast \tau) & = -d(\alpha \wedge f^\ast \tau) + d\alpha \wedge f^\ast \tau \\ & = -d(\alpha \wedge f^\ast \tau) + f^\ast (\omega \wedge \tau) \\ & = -d(\alpha \wedge f^\ast \tau), \end{align*} where once again $f^\ast (\omega \wedge \tau) = 0$ since $\omega \wedge \tau$ is an $(n+1)$-form on $S^n$, we get that $\alpha' \wedge d\alpha' = \alpha \wedge d\alpha + d(-\alpha \wedge f^\ast \tau + \eta \wedge (d\alpha + f^\ast d\tau)).$ So by Stokes' theorem, \begin{align*} \int_{S^{2n-1}} \alpha' \wedge d\alpha' & = \int_{S^{2n-1}} (\alpha \wedge d\alpha + d(-\alpha \wedge f^\ast \tau + \eta \wedge (d\alpha + f^\ast d\tau))) \\ & = \int_{S^{2n-1}} \alpha \wedge d\alpha, \end{align*} showing that the integral is independent of the choices.
Remark: The number $H(f) = \int_{S^{2n-1}} \alpha \wedge d\alpha$ is called the Hopf invariant of the map $f$. Here we showed that it only depends on $f$. You can also show that it only depends on the homotopy class of $f$, which isn't too difficult.