(Note: Using Einstein summation convention throughout)
I'm trying to understand Theorem 2.2.1 in Wald's General Relativity book. We have a smooth manifold $M$, with $p\in M$ and $(O,\psi)$ an open chart containing it. Let $f\in C^{\infty}(M)$ and $X^i(f)\equiv\frac{\partial}{\partial x^i}(f\circ\psi^{-1})\big|_{\psi(p)}$.
There's a result that if $F:\mathbb{R}^n\to\mathbb{R}$ is smooth, then for each $a=(a^1,\ldots,a^n)\in\mathbb{R}^n$, $$F(x)=F(a)+(x^i-a^i)H_i(x)\tag{1}$$ where $H_i(x)=\frac{\partial F}{\partial x^i}\big|_{x=a}$. So now letting $F=f\circ\psi^{-1}$ and $a=\psi(p)$, for all $q\in O$, we get $$f(q)=f(p)+[x^i\circ\psi(q)-x^i\circ\psi(p)]H_i(\psi(q))\tag{2}$$
So far it's clear. The next step is what I don't understand:
Let $v\in T_p(M). [...]\ $we apply $v$ to $f$. Using equation (2), the linearity and Leibnitz properties of $v$, and the fact that $v$ applied to a constant (such as $f(p)$) vanishes, we obtain $$v(f)=v[f(p)]+[x^i\circ\psi(q)-x^i\circ\psi(p)]\ \bigg|_{q=p}v(H_i\circ\psi)+(H_i\circ\psi)\big|_pv[x^i\circ\psi-x^i\circ\psi(p)]\tag{3} \\=\big(H_i\circ\psi(p)\big)\ v(x^i\circ\psi)$$
There's a lot I don't get - Why is the LHS not $v[f(q)]$? And in the very last bracket on RHS, why is it $[x^i\circ\psi-x^i\circ\psi(p)]$ and not $[x^i\circ\psi(q)-x^i\circ\psi(p)]$?
Sorry if this seems like a naive question - I guess my multivariable calculus is rusty..
The author appears to follow the convention of not including the argument of a function when a vector is being applied to it. Thus $v(f)$ is not written as $v[f(q)]$ so as to more easily distinguish it from the case $v[ f(p) ]$ where $p$ is now a constant. This explains what is happening on the LHS and RHS.