Let $M$ be a manifold and $X$ a vector field on $M$.
There are three ways how one can think of this vector field:
As a (smooth) section of the tangent bundle:
$X_s: M \rightarrow TM$
As a derivation:
$X_d: \mathcal{C}^{\infty}(M) \rightarrow \mathcal{C}^{\infty}(M)$
As a $\mathcal{C}^{\infty}(M)$-linear map:
$X_l: \Omega^1(M) \rightarrow \mathcal{C}^{\infty}(M)$
Identifying the tangent space at a point $p \in M$ as the vector space of derivations at that point, i.e. $T_pM \simeq \{ \partial_p :\mathcal{C}^{\infty}(M) \rightarrow \mathbb{R} \, | \, \partial_p \text{ derivation}\}$, I am aware of the following identification: $$X_s(p)(f) = X_d(f)(p) \in \mathbb{R} \, , $$ for $p \in M$ and $f \in \mathcal{C}^{\infty}(M)$.
Now my question is, how does the identification of $X_l$ fit into this picture? More precisely:
Given, $p,q \in M$, $\mathcal{C}^{\infty}(M)$ and $\omega \in \Omega^1(M)$ such that $$X_l(\omega)(q) = X_d(f)(p)\, ,$$ holds. How are $p,f,\omega$ and $q$ related?
Apparently, for the exterior derivative $\omega = df$ we have $$X(\omega) = \omega(X) = df(X) = X(f).$$ So, the answer is $\omega= df$ and $p=q$.