According to the discussion in this thread, even though the spherical coordinate $\theta$ on $S^1$ is a local function, $d\theta$ is a global form on $S^1$. How to prove this rigorously? The discussion also points out it is the pullback of $\frac{-ydx+xdy}{x^2+y^2}$. Why?
My attempt:
For the first question: as hinted I probably want to show that two local form $d\theta$, and $d\theta+\pi$ agrees on their overlaps. The point is that I want to evaluate $d(\theta+\pi)(\frac{\partial }{\partial\theta})$. To do that take a function $f: \mathbb R \to \mathbb R$, $d(\theta+\pi)(\frac{\partial }{\partial\theta})(f) = \frac{\partial}{\partial\theta}(f \circ (\theta+\pi))$, where $f\circ(\theta + \pi)(e^{i\theta}) = f(\theta+\pi)$. So $d(\theta+\pi)(\frac{\partial }{\partial\theta})(f) = f'|_{\theta+\pi}$ as desired. Then $d(\theta+\pi)$ and $d\theta$ coincides on their overlap, thus we can extend $d\theta$ to a global form on $S^1$.
For the second question: Let $i: s^1 \to \mathbb R^2$ be the embedding. Then $i^{\ast}(\frac{-ydx+xdy}{x^2+y^2}) = \frac{-y\circ i(dx \circ i)+x\circ i(dy \circ i)}{(x\circ i)^2+(y\circ i)^2}$. The key observation is that in spherical coordinate $x\circ i(\theta) = cos\theta, y \circ i(\theta) = sin\theta$.
Am I right in my reasonings?