Let $M$ be a n manifold with boundary and $p\in \partial{M}$.
$v\in T_pM\backslash T_p\partial{M}$ is inward pointing if there exists some $\epsilon>0$ and a smooth curve $c:[0,\epsilon)\rightarrow M$ such that $c(0)=p$ and $c((0,\epsilon))\subseteq intM$ and $c'(0)=v$.
Problem:
Let $M$ be a smooth n manifold and $p\in \partial{M}$. Let $(U,x^1,.....,x^n)$ be a chart about $p$. $v=\sum_{i=1}^nv^i\frac{\partial}{\partial{x^i}} \in T_pM$ is inward pointing if and only if $v^n>0$.
My attempt:
Suppose $v\in T_pM$ is inward pointing. This means $v\notin T_p\partial{M}$ and there exists some $\epsilon>0$ and a smooth curve $c:[0,\epsilon)\rightarrow M$ such that $c(0)=p,$ $c((0,\epsilon))\subseteq intM$ and $c'(0)=v$. By definition of $\mathbb{H}^n$, we have that on $U$, $x^n\geq 0$. Also, it is easy to show that $v^n=\frac{d}{dt}|_{t=0}(x^n\circ c)$. Also note that we may assume $x^n>0$; for otherwise, $v\in T_p\partial{M}$.
However, I'm not sure what else to say. If $(U,\phi)$ was centered at $p$, then I can prove the result, but in this case, the chart may not necessarily be centered at p.,
Note, $(U,\phi)=(U,x^1,....,x^n)$ is a chart in $M$. Thus $\phi:U\rightarrow \phi(U)$ is a diffeomorphism. Hence $\phi(p)=\phi(c(0))\in \partial{\mathbb{H}}^n$. So $x^n(c(0))=0$. Thus $v^n=\frac{d}{dt}|_{t=0}(x^n\circ c)=\lim_{t\rightarrow 0^{+}}\frac{x^n(c(t))}{t}\geq 0$. We cannot have the latter expression equal to $0$, for as you said, this would imply that $v\in T_p(\partial{M})$. Therefore we must have $v^n>0$. Conversely, if $v^n>0$, define for sufficiently small $\epsilon>0$, $\gamma: (-\epsilon,\epsilon)\rightarrow M$ by $\gamma(t)=\phi^{-1}(v^1t,.....,v^nt)$. We must show that $\gamma((0,\epsilon))\subseteq intM$. Indeed, for $t'\in (0,\epsilon)$, $\gamma(t')\in U$. Thus, $(v^1t,.......,v^nt)\in \phi(U)$ , where $v^nt>0$, and so $\gamma(t')\in intM$. $\gamma$ is the curve we seek. $\square$