Suppose $x=x(t)$. Let $dx$ and $dt$ be 1-forms.
Does there exist a rigorous theorem for converting $dx=f(t)dt$ into the derivative ("fraction") expression $\frac{dx}{dt}=f(t)$?
Is this possible in general for other possibly multivariate functions (or does it require that $x$ is only a function of $t$). Or is this just false?
My guess is that by the linear independence of the basis differential form dt, it is possible to expand in the following manner and simply equate coefficients of $dt$.
$dx=\frac{dx}{dt}dt$
Therefore, $\frac{dx}{dt}=f(t)$.
As I commented, you want to think of a mapping $x\colon\Bbb R\to\Bbb R$. Using $t$ as a coordinate on the domain and $u$ on the range, you pull back $du$ by the mapping $u=x(t)$. Then you get $dx = x^*du = x'(t)\,dt$. Setting this equal to $f(t)\,dt$, you have $x'(t)\,dt = f(t)\,dt$, whence $x'(t)=f(t)$.
You asked about the multivariate situation. Consider, for example, $$dx = P(u,v)\,du + Q(u,v)\,dv.$$ Then this reduces likewise to the equations $$\frac{\partial x}{\partial u} = P(u,v) \quad\text{and}\quad \frac{\partial x}{\partial v} = Q(u,v).$$ Can you make the analogous argument?