Here is the version of proof that I am trying to understand, along Moser’s method.
Towards the end of 2nd page we find $\omega _t$ restricted to $N$ being non degenerate, which is fine. What I am unable to understand is the very next line stating that this implies $\omega _t$ on a neighborhood of $N$ too is non degenerate for all $t$. Is this something trivial as in is it a property of differential forms (if so how)? Or is it a corollary of tubular neighborhood theorem (which I am not very familiar with)?
This is mostly point-set topology. Because this is a local discussion I will work in a chart $\Bbb R^{2n} \hookrightarrow M$. The set $N$ is not important: what we will show is that if $\omega_p$ is nondegenerate then so is $\omega_q$ for $q$ sufficiently near $p$. That is, the set of $p$ for which $\omega_p$ is nondegenerate is open. Since $\omega_p$ is nondegenerate for all $p \in N$, the set of $p$ for which it's nondegenerate is an open set containing $N$ as desired.
The point is that $\omega$ gives a smooth map $\Bbb R^{2n} \to \Lambda^2(\Bbb R^{2n})$ sending $p \mapsto \omega_p$. Then there is a smooth map $\Lambda^2(\Bbb R^{2n})^* \to \Lambda^{2n}(\Bbb R^{2n})^* \cong \Bbb R$, where this map is $\omega \mapsto \omega^n$ and the last isomorphism is the determinant.
$\omega_p$ is nondegenerate if and only if $\det \omega^n \neq 0$. Because $p \mapsto \det \omega^n_p$ is continuous, and $\Bbb R \setminus 0$ is open, it follows that $$\{p \mid \omega_p \text{ is nondegenerate}\}$$ is open.