I have an issue with the following proof. Here, $f_a$ is defined only on $U_a$. So to make the function $f=\sum_a \psi_a f_a$ defined on $M$, we need to extend $\psi_a f_a$ to $M$. We do this (as in Lemma 2.26) by making it $0$ on $M \backslash supp \psi_a$. However, note that we have not extended $f_a$ to $M$.
This is what I assume is done in the below proof. But then my problem is that we cannot make sense of $f_a(p)d\psi_a|_p(v)$ in the sum. I can see that since $p$ is a boundary point, it only belongs to a boundary chart so we must have $f_a(p)=0$. However, what happens if $U_a$ is an interior chart? Then how do we make sense of $f_a(p)$ below? So how can we apply the product rule when we have not extended $f_a$ to the whole of $M$?
I am really confused with this because for $U_a$ interior, we have $f_a(p)=1$, so I don't get why the first term in parentheses is $0$.
To consider the differential of $f$ at $p$, we need to look at $f$ in a neighborhood of $f$ that intersects finitely many $\supp \psi_a$. How do we ensure that the $\psi_a$'s here are the ones subordinate to the boundary charts?
Also how do we know that there is at least one $\alpha $ for which the second term is positive?
