Let $M$ be an $1$-dimensional manifold. Consider a 1-form $w\in\Omega^1(M)$. I saw that we can write it $$w(t)=g(t)\,dt$$ but I have a question about this:
$w(t)\in \bigwedge^1(T^*M)$, so we can write it in a basis of $\bigwedge^1(T^*M)$. $dt\in \Omega^1(M)$ spans $\Omega^1(M)$, not $\bigwedge^1(T^*M)$, while $dt(t)$ does span $\bigwedge^1(T^*M)$. So, shouldn't we write $w(t)=g(t)\,dt(t)$ instead? This may be nitpicking, but I want to make sure I am understanding this correctly.
Let $\omega\in \Omega^{1}(M)$. For $t\in M$, we have $\omega(t)\in \Lambda^1 T_t^\ast M$. Now take a chart $\varphi\colon U\cong V$ with $t\in U$. Then $\frac{\partial}{\partial{t}} \in T_tM=\operatorname{Der}(\mathscr{C}^\infty_t, \mathbb{R})$ defined by $\frac{\partial{}}{\partial{t}}([f])=\frac{\partial{f \circ \varphi^{-1}}}{\partial{x}}(\varphi(t))$ ($\partial{}/\partial{x}$ denotes the usual (partial) derivative) gives you a basis of $T_tM$. Thus, the dual element $dt$ will give you a basis of $T^\ast_t M$. Hence, you may uniquely write $\omega(t)=g(t)dt$, where $g(t)\in \mathbb{R}$.
Since this works for every element $t\in U$, you may $\textit{locally}$ write $\omega$ as $\omega(t)=g(t)dt$. To be precise, $\omega\vert_U = g(t)dt$, where $g\colon U\to \mathbb{R}$ is the function assigning each point $t$ the unique coefficient $g(t)$ satisfying $\omega(t)=g(t)dt$.
And strictly speaking, yes, $dt$ depends on the chosen point $t$ as you can tell from the definition of $\frac{\partial{}}{\partial{t}}$; so it would be perfectly correct to write $\omega(p)=g(p)dt(p)$ for $p\in U$.