Let $M\subset \mathbb{R}^n$ be an $m$-dimensional manifold (in the ordinary Euclidean sense).
Given a point $p\in M$ we define a derivation $D_p$ (at $p$) to be linear functional on the space of smooth functions of $M$ such that $D_p(fg) = f(p)D_p(g) + g(p)D_p(f)$.
For any $p\in M$, there is an open set $U_p\subset \mathbb{R}^n$ containing $p$, and an open set $V_p\subset \mathbb{R}^m$ containing $0$, together with a diffeomorphism $\varphi_p:V_p \to (U_p\cap M)$. We define $\frac{\partial}{\partial x_i}_p(f) = \partial_i (f\circ \varphi)(0)$. It is clear that $\frac{\partial}{\partial x_i}_p$ is a derivation at $p$. Note, this definition depends on the choice of a chart for $M$.
Fix $1\leq i\leq m$. Let us define a vector field $V$ on a manifold $M$ by, $$ V(p) = \frac{\partial}{\partial x_i}_p $$ Where we make a pointwise choice of atlas for each $p$. Why is it true that $V(f)$ is a smooth function provided that $f$ is smooth?
(Note, in books vector fields are defined locally based on a chart instead of punctually. Wondering whether this one single consistent chart is really necessary to have a smooth vector field. )