Let $X$ be a manifold, for simplicity, 1-dimensional. I want to understand what this means:
"Locally a differential form $s$ is of the form $f(x)dx$".
As I understand it, differential forms are sections of $T^*X$, thus a map $X \to T^*X$. Taking a local trivialization $U$ of $X$, $s$ can be represented by function $g:U \to X\times \mathbb R$. I think $g$ should be $f$ and the $dx$ simply represents that $s(x) ( \frac d {dx}) = g(x)$.
On the other hand, if we have a chart $\phi:U\to \mathbb R^n$ with $\phi(x)=0$, we can consider the pull-back $\phi^* s$, a section on $T^*\mathbb R$. On $T^* \mathbb R$, $dx$ is well defined, so $\phi^*(s)$ is of the form $h(x)dx$. Is it then true, that $\phi^*|_U(s)=f(x) dx$?
I think this should be true, because $\phi^*(s)(0)=s(\phi(x))=s(x)$.
I am not sure. Does this make sense?
Edit: So far an answer has been posted, but it doesn't really deal with my thoughts on the topic and what i need clarification of. Any additional answers would be appreciated.
Edit2: Still not answered!
I think you make simple things complicated. In your chart $x: U\to \bf R$ is a function. Thus $dx$, its differential, is a nowhere vanishing differential form, and as your manifold is one dimensional, in this chart every form is proportional to $dx$, ie can be written as $\ f(x) dx$.