I need help to understand a proof that there is at most one operator $*$ satisfying $\eta\wedge*\zeta=\langle\eta,\zeta\rangle\omega$ This is from some lecture notes I found online (LMU Munich):

If $*_1,*_2$ are two operators satisfying the defining property of the Hodge star operator, then $\eta\wedge(*_1\zeta-*_2\zeta)=0$. If the second factor is non-zero for some $\zeta$,
then one can easily find $\eta$ so that the product is a mulitple of the volume form $\omega$.
This proves uniqueness.

According to the lecture notes, this is also valid for indefinite metrics.
Question: Can someone explain the bold sentence? What $\eta$ is the author talking about?
My thoughts:
As far as I understand, this is a proof by contradiction: Assuming $*_1\zeta\neq *_2\zeta$, we can find $\eta$ such that $\eta\wedge(*_1\zeta-*_2\zeta)\neq 0$, but because of $\eta\wedge*_1\zeta=\langle\omega,\eta\rangle\omega=\eta\wedge*_2\zeta$ we should have $\eta\wedge(*_1\zeta-*_2\zeta)=\eta\wedge*_1\zeta-\eta\wedge*_2\zeta=0$.
My first guess was $\eta=\zeta$, but this doesn't work. My second guess was $\eta=*_1\zeta-*_2\zeta$, but this didn't work either. (Or maybe they do work and I just don't get it.)
 A: Suppose $\alpha$ is a $k$-form and define the application $g_{\alpha} : \Lambda^{\dim M-k}T^*M \to \Lambda^{\dim M} T^*M$ defined by $f_{\alpha}(\beta) = \alpha\wedge \beta$.
Your question can be rewritten as: why $g_{\alpha}$ is not the null-application if $\alpha \neq 0$?
This is a local question: in local coordinates around $p\in M$, $\alpha = \sum_{i_1<\cdots < i_k} \alpha_{i_1,\ldots,i_k} \mathrm{d}x^{i_1}\wedge\cdots\wedge \mathrm{d}x^{i_k}$. If $\alpha$ is non-vanishing, then there exists $i_1<\ldots<i_k$ such that $\alpha_{i_1,\ldots,i_k}\neq 0$ on a neighbourhood of $p$. Without loss of generallity (by re-ordering the coordinates), suppose $(i_1,\ldots,i_k) = (1,\ldots,k)$. Then,
\begin{align}
\alpha\wedge \left(\mathrm{d}x^{k+1}\wedge \cdots \wedge \mathrm{d}x^{\dim M}\right) = \alpha_{1,\ldots,k}\mathrm{d}x^1\wedge\cdots\wedge \mathrm{d}x^{\dim M} \neq 0
\end{align}
We have then found a local $\beta$ such that $\alpha \wedge \beta$ is non vanishing in a neighbourhood of $p$. An argument of partition of unity show that $\beta$ can be defined globally (just extend it with $0$), and there exists $\beta$ such that $\alpha \wedge \beta$ is not the zero $\dim M$-form.
This shows that if $\eta \wedge (*_1\zeta - *_1\zeta) = 0$ for every $k$-form $\eta$, then $*_1\zeta - *_2\zeta$ the zero $(\dim M -k)$ form.
Comment: the fact that $M$ is orientable is not required to show this property, that is the uniqueness of the Hodge operator. But it is required to show its existence!
Edit: here is how to extend globally $\beta$. Suppose $\alpha$ is non zero at $p$. Suppose $U$ is an open subset containing $p$, and on which we have the above coordinates. Let $\psi : M \to \mathbb{R}$ be a smooth function with compact support, and with $\mathrm{supp}\psi \subset U$ and $\psi(p) = 1$. Then, the differential form $\beta$ defined to be $0$ outside of $U$ and $\psi \cdot \mathrm{d}x^{k+1}\wedge \cdots \wedge \mathrm{d}x^{\dim M}$ in $U$ (in the above coordinates) is well-defined, smooth, is non zero, and $\alpha \wedge \beta$ is a non-zero $\dim M$-form, because $\alpha\wedge\beta (p) \neq 0$.
