In Lee's Introduction to Smooth Manifolds, Chapter 8 (page 176) it's written that given a manifold $M$ and arbitrary subset $A \subseteq M$, $X$ is said to be a smooth vector field along $A$ if for each point $p \in A$, there is a neighborhood $V$ of $p$ in $M$ and a smooth vector field $\tilde{X}$ on $V$ that agrees with $X$ on $V \cap A$.
Just to be certain, this is saying that $V$ is an open neighborhood and $\tilde{X}$ being smooth on $V$ means each of the component functions is smooth right? Thank you!
Im Lee's book neighborhoods are always open sets. See Appendix A p.596 :
A vector field $X$ on a smooth manifold $M$ is a smooth section of $\pi : TM \to M$. This is a coordinate free definition. However, if we introduce coordinates based on a smooth coordinate chart $(U,(x^i))$ on $M$, we can represent $X \mid_U$ via its component functions in this chart. These must of course be smooth. See Proposition 8.1 (Smoothness Criterion for Vector Fields).
In particular $\tilde X$ is a smooth section of $\pi_V : TV \to V$ and no component functions are needed here. But of course we can introduce coordinates and get component functions which have to be smooth. W.l.o.g. we may even assume that $V$ is the domain of a smooth coordinate chart.