This is just here so that I can mark the question as answered for people looking for it. The proof provided above should no longer have any errors, but if someone finds a better way of doing things (or an error), please post a new response, and I'll mark that as the preferred answer over this one. Thank you!
I didn’t fully read it, but yes that’s essentially it. Given a $k$-form $\omega$ such that $\langle \omega,\cdot\rangle_{L^2}=0$, you fix a point $p\in M$, and a sufficiently small open neighborhood $U$ of $p$, on which we have a pointwise ‘orthonormal’ frame of vector fields, which gives rise to 1-forms which gives rise to ‘orthonormal’ k-forms $\{\alpha^{i_1}\wedge\cdots\wedge \alpha^{i_k}\}$. Note that verifying that the wedges of the 1-forms are ‘orthonormal’ with respect to the induced pseudo-inner product on $\Lambda^k(T^*M)$ takes a bit more work in the non-Riemannian case… you can’t just say that oh because it’s a subspace of $T^0_k(TM)$ the restriction is non-degenerate (e.g in 2D Minkowski, restricting to the 45 degree lines gives the zero tensor which is as far as you can get from non-degeneracy). In any case, this is a linear algebra fact so do it on one vector space first.
From here, it’s a matter of reducing to the case of the classical ‘fundamental lemma of calculus of variations’. Fix an arbitrary smooth function $f:M\to\Bbb{R}$ with support that is compact and contained in $U$, and fix an increasing multindex $I=(i_1,\dots, i_k)$, and consider the $k$-form $f\alpha^I\equiv f\alpha^{i_1}\wedge\cdots\wedge \alpha^{i_k}$. Ok, very strictly speaking, I’m abusing notation slightly here, because $f$ is defined on all of $M$, while $\alpha^I$ is only defined on $U$, so the product is a-priori only defined on $U$; but note that since $f$ has compact support, I can define it to be $0$ outside $U$, and the result is a smooth $k$-form on $M$. Now, we compute:
\begin{align}
0&=\langle\omega,f\alpha^I\rangle_{L^2}
=\int_M\pm\omega_If\,dV_g
=\int_U\pm\omega_If\,dV_g,
\end{align}
where the first equality is by assumption, the second by ‘orthonormality’ (the $\pm$ sign is given by $\langle\alpha^I,\alpha^I\rangle_{\bigwedge^k(T^*M)}$ (a constant function with value $1$ or $-1$ due to ‘orthonormality’)), and the last equal sign is because $f$ has support in $U$ . Since $f$ is an arbitrary smooth function with compact support in $U$, this is back to the classical situation, so we conclude that $\omega_I=0$. Since the multindex $I$ was arbitrary, it follows $\omega=0$ on $U$. Finally, since $p$ was arbitrary, it follows $\omega=0$ on $M$, proving the non-degeneracy.
What makes your proof slightly unpleasant to read is that you’re doing too many things simultaneously. For instance, you’re defining bump functions explicitly, and you’re trying to reprove the classical fundamental lemma of calculus of variations (this is a standard result, so if you want, prove that first, and then just invoke it here, but don’t do one proof inside of another inside of another). In fact, you can drop the smoothness assumptions and show that if $\omega$ is an $L^2$ differential $k$-form on $M$ such that $\langle\omega,\cdot\rangle_{L^2}=0$, then $\omega=0$ a.e on $M$ (with respect to the measure $dV_g$); the proof is virtually unchanged.
Best Answer
The trick is that you can actually work globally instead of locally; all the local-to-global work can be done just on $M$ and $N$ themselves using their orientability. So, just let $\omega$ and $\eta$ be any nowhere vanishing top forms on $M$ and $N$ (these exist since $M$ and $N$ are orientable), so $\pi_1^*\omega\wedge \pi_2^*\eta$ is a nowhere vanishing top form on $M\times N$. Any other top form is then a scalar multiple of $\pi_1^*\omega\wedge \pi_2^*\eta$ at each point, and thus can be written in the form $h\pi_1^*\omega\wedge \pi_2^*\eta$ for some smooth function $h$.
(If you want $\omega$ and $\eta$ to be compactly supported as well, then you can choose them to have large enough supports so that $\pi_1^*\omega\wedge \pi_2^*\eta$ is still nonzero on the entire support of $\chi$.)