Suppose $M$ is an $n$-dimensional manifold with boundary $\partial M.$ Consider a point $p\in \partial M,$ an open set $U\subseteq M$ containing $p,$ local coordinates $(x^i)_{1\le i\le n},$ and the basis $\left(\frac{\partial}{\partial x^i}\right)_{1\le i\le n}$ for $T_pM$. Then, it is intuitively obvious (drawing pictures) that "$v=\sum^n_{i=1} a_i\frac{\partial}{\partial x^i}\in T_p(\partial M)$ if and only if $a_n=0.$" I have seen this several times stated without proof, and it seems sloppy to me. I'd like to offer my thoughts on this and perhaps see a more direct proof because I think I am making this too difficult.
In the first place, $T_pM$ and $T_p(\partial M)$ are technically two different spaces, so to make sense of this, I am supposing that $T_p(\partial M)\subseteq T_pM$ via the inclusion $i:\partial M\to M, $ so that for smooth functions $f: M\to \mathbb R$, if $v\in T_p(\partial M)$, it is identified with the vector $i_*v$, which satisfies $i_*v(f)=v(f\circ i).$ Then, $i_*v=\sum^n_{i=1} a_i\frac{\partial}{\partial x^i}$ so $i_*v(x_n)=a_n$ and also $i_*v(x_n)=v(x_n\circ i)=0$ because $x_n\circ i=0$. Therefore, $a_n=0,$ as desired.
Or, one can take $F:U\to \mathbb R:(x^1,\cdots, x^n)\mapsto x^n$ so that $F$ is a defining function for $\partial M$ and therefore $T_p \partial M=\text{ker}\ dF$, from which the result follows, but this seems like too big a gun for the purpose.