I'm reading Semi-Riemannian Geometry by Newman - the following theorem
Let $M$ be a smooth manifold, and let $X$ be a vector field on $M$. Then $X\in \mathfrak{X}(M)$ iff for every chart $(U,(x^i))$ on $M$, the components of $X$ are in $C^{\infty}(U)$.
The proof for $\implies$ is given as follows:
Let $X=\sum_i\alpha^i(\partial/\partial x^i)$ be the local coordinate expression of $X$ w.r.t. $(U,(x^i))$, and let $p\in U$. Since $x^i\in C^{\infty}(U)$, (by some previous theorem) there is a function $\tilde x^i\in C^{\infty}(M)$ and a neighborhood $\tilde U\subseteq U$ of $p$ such that $\tilde x^i$ and $x^i$ agree on $\tilde U$. So $$X(\tilde x^i)=\bigg(\sum_j\alpha^j\frac{\partial}{\partial x^j}\bigg)(\tilde x^i)=\sum_j\alpha^j\frac{\partial\tilde x^i}{\partial x^j}=\alpha^i$$ on $\tilde U$. Since $X\in\mathfrak{X}(M)$ and $\tilde x^i\in C^{\infty}(M)$, $X(\tilde x^i)\in C^{\infty}(M)$. So $\alpha^i$ is smooth on $\tilde U$. Since $p$ was arbitrary, $\alpha^i$ is smooth on $U$.
Now a previous result states that the local coordinate expression for a $v\in T_p(M)$ w.r.t. a chart $(U, (x^i))$ covering $p$ is $$v=\sum_i v(x^i)\frac{\partial}{\partial x^i}\bigg|_p$$
I can translate this to $$X_p=\sum_i X_p(x^i)\frac{\partial}{\partial x^i}\bigg|_p=\sum_i X(x^i)(p)\frac{\partial}{\partial x^i}\bigg|_p=\bigg(\sum_i X(x^i)\frac{\partial}{\partial x^i}\bigg)(p) \\\implies X=\sum_i X(x^i)\frac{\partial}{\partial x^i}$$
So basically the local coordinates of a smooth vector field $X$ w.r.t. a chart $(U,(x^i))$ are $X(x^i)$. I'm trying to understand why the proof went through the extra step of considering another chart $(\tilde U, (\tilde x^i))$. Do we run into problems when trying to prove that $X(x^i)$ is smooth? Since both $X$ and $x^i$ are also smooth, it's not immediately obvious to me why we'd have a problem.