Check that the $1$-form $d\,\text{arg}$ in $\mathbb{R}^2 - \{0\}$ is just the form$${{-y}\over{x^2 + y^2}}\,dx + {{x}\over{x^2 + y^2}}\,dy.$$
My solution is as follows.
Observe that we can define $\text{arg}\,z= \tan^{-1}(y/x)$ locally. This definition only works if $\text{arg}\,z$ is congruent to $\theta$ modulo $2\pi$ for some $\theta \in (-\pi/2,\pi/2)$. If $\text{arg}\,z$ is outside this range, we just add $\pi$ to it, leaving the exterior derivative unchanged. Taking this to be a $0$-form, we calculate its exterior derivative, as follows.$$d\left(\tan^{-1}\left({y\over{x}}\right)\right) = {{\partial f}\over{\partial x}}dx + {{\partial f}\over{\partial y}}dy = -{y\over{x^2 + y^2}}dx + {x\over{x^2 + y^2}}dy.$$In neighborhoods of $\pm\pi/2$, we instead define $\text{arg}\,z = \cot^{-1}(x/y)$, seeing that it is not clear that our earlier definition of $\text{arg}\,z$ is even continuous near $\pm\pi/2$. $\cot^{-1}(x/y)$ gets us the same exterior derivative as $\tan^{-1}(y/x)$, and it is well-defined, so we are done.
My question is, is what I have valid? And is there a cleaner/simpler way of doing the problem/alternate perspective I'm missing? Much thanks in advance.
Briefly, your approach is good, and I know of no simpler way to discover the formula for the $1$-form $\omega := d\arg$.
Depending on the level of rigor expected, you might want to prove that $\tan^{-1}(y/x) = \cot^{-1}(x/y)$ in the first quadrant, and that $\tan^{-1}(y/x) = \cot^{-1}(x/y) - \pi$ in the fourth quadrant. That is, if $\tan^{-1}$ and $\cot^{-1}$ are principal branches, the formula $$ \arg(x + iy) = \begin{cases} \cot^{-1}(x/y) - \pi & y < 0, \\ \tan^{-1}(y/x) & x > 0, \\ \cot^{-1}(x/y) & y > 0 \end{cases} $$ defines a smooth branch of $\arg$ on the cut plane $\mathbf{C} \setminus (-\infty, 0]$, even though the respective domains overlap.
Alternatively, since you have the formula for $\omega$, it's faster to prove that if $P$ is the polar coordinates map $P(r, \theta) = (r\cos\theta, r\sin\theta)$ (or, if you prefer, the exponential map $P(r, \theta) = e^{r} \cos\theta, e^{r} \sin\theta)$), then $P^{*}\omega = d\theta$.