R-linear functionals on manifolds Surely the following is well known:
Let $X$ be a (differentiable) manifold, 
$R$ the ring of continuous/smooth real functions on $X$, 
$V$ the $R$-module of all continuous/smooth vector fields on $X$, and $\operatorname{Hom}_R(V,R)$ the $R$-module of all $R$-linear maps from $V$ to $R$.
Using 'local bump functions', it's not difficult to see that whenever $F$ is such an $R$-linear map, and whenever $f, g$ in $V$ and $f = g$ locally around some point $x$, then also $F(f) = F(g)$ locally around $x$.
Now I had expected something stronger to hold, namely that $F(f)(x)$ depends only on $f(x)$, i.e. for any $x$ in $X$, if $f(x) = g(x)$, then $F(f)(x) = F(g)(x)$.
But I can't find a proof, nor a counterexample.
Anyone?
(Note that derivatives are not $R$-linear.)
 A: Unless I'm missing something, your "strengthened" version is correct as well.
If you don't mind, I'm going to use notation which is more familiar to me.  So, I'll let $X$ denote a vector field.  Since you already have shown that if $X = Y$ on an open subset $U\subseteq M$, that $F(X)(p) = F(Y)(p)$ for all $p\in U$, we may reduce your question to a local one.  That is, we may assume without loss of generality, that $M = \mathbb{R}^n$ (or, alternatively, just work on $U$, assuming $U$ is a chart.)
Thus, we may express $X = \sum a_i \frac{\partial}{\partial x^i}$ for some functions $a_i$ defined on $U$.  Let $Y = \sum b_i \frac{\partial}{\partial x^i}$ be any vector field and assume $a_i(p) = b_i(p)$ - that is, $X(p) = Y(p)$.  I claim that $F(X)(p) = F(Y)(p)$, so that, in particular $F(X)(p)$ only depends on $X(p)$.
To see this, just use $R$-linearity:  \begin{align*} F(X)(p) &= F\left(\sum a_i \frac{\partial}{\partial x^i}\right)\\ &= \sum a_i(p) F\left(\frac{\partial}{\partial x^i}\right)(p) \\ &= \sum b_i(p)F\left(\frac{\partial}{\partial x^i} \right)(p)\\ &=F\left(\sum b_i \frac{\partial}{\partial x^i}\right)\\ &=F(Y)(p).\end{align*}
