The fact I'm trying to prove is that every bilinear form $\omega$ on a set $\omega\subseteq\mathbb{R}^2$ can be written in the form $\omega = f\,dx\wedge dy$, where $f: \Omega\rightarrow\mathbb{R}$ is a function.
I know that if $L$ is an skew-symmetric bilinear form, and for some $a\in\mathbb{R}^2$: $$L((x_1, y_1), (x_2, y_2)) = L(e_1, e_2)(x_1y_2- x_2y_1) = L(e_1, e_2)(dx\wedge dy)_a(x, y)$$
I assume this is close to what I want? Would the $f$ in my fact just be equal to $L(e_1, e_2)$? However, this proof relies on $L$ being skew-symmetric (or at least, symmetric), an additional requirement that is not present on $\omega$ in my original fact.
Separately, is this fact directly analogous to the second, and more general, fact that every bilinear form $\omega$ on $\Omega\subseteq\mathbb{R}^n$ can be written as: $$\omega = \sum_{1\le i<j\le n}\omega_a(e_i, e_j)\, dx\wedge dy$$