If $\omega$ is a volume form, then $X\mapsto i_X\omega$ generates all $(n-1)$-forms

This is mostly a pointwise statement. That $\omega$ is a volume form means that $\omega_p(X_1,\cdots,X_n) \neq 0$ if and only if $X_1,\cdots,X_n$ is a basis of $T_pM$. ($\omega_p$ is a non-zero "determinant".)

The map \begin{align*} \iota_p:T_pM &\to \Lambda^{n-1}_p(T_pM) \\ X &\mapsto \iota_X\omega \end{align*} is then injective, since $\omega(X_1,\cdot,\cdot,\cdots)=\omega(X_2,\cdot,\cdot,\cdots)$ implies $\omega(X_1-X_2,\cdot,\cdot,\cdots)=0$, and thus $X_1-X_2$ can't be completed to a basis. Therefore, it must be zero. By dimensional reasons, the map is then an isomorphism.

There is still the issue of smoothness when we go from the pointwise statement to the differential one. By taking local charts and trivializations, if $\omega=gdx_1\cdots dx_n$ and we are taking a differential form $\eta=\sum f_i dx_1 \cdots \widehat{dx_i}\cdots dx_n$, we are getting $X=\sum_i (-1)^{i+1}\frac{f_i}{g}\partial_i$ as our output, which is smooth.