Setup: Let $$\mathbb{R}^n\_ := \{(x^1,\ldots,x^n) \subset \mathbb{R}^n: x^1 \leq 0\}$$ and $$\partial \mathbb{R}^n\_ := \{0\} \times \mathbb{R}^{n-1}$$ i.e. $$x \in \partial \mathbb{R}^n\_$$ are on the form $$x = (0,x^2,\ldots,x^n).$$
Now, let $p \in M$ be such that $x^1(p) = 0$ for a chart $(U,\varphi = (x^1,\ldots,x^n))$. Then we have that for any other chart $(V,\psi = (y^1,\ldots,y^n))$ with $p \in V$,we get $y^1(p) = 0$.
Proof: Suppose $y^1(p) < 0$. Let $v \in \mathbb{R}^n$ be such that $$D(\varphi \circ \psi^{-1})_{\varphi(p)}(v) = e_1 = (1,\ldots,0).$$ This step is possible, since both $\varphi$ and $\psi^{-1}$ are diffeomorphisms, hence $\varphi \circ \psi^{-1}$ is a diffeomorphism, hence $D(\varphi \circ \psi^{-1})$ is a linear isomorphism, i.e. surjective.
Now, for small $t$, set $c:I \to \mathbb{R}^n$ so that $c(t) = \psi(p)+tv$. We can then choose $t$ small enough so that $c(t) \in \psi(U \cap V)$. So, for small $t$ (pick some open interval $I_\epsilon = (-\epsilon,\epsilon)$), we have that $c:I_\epsilon \to \psi(U \cap V) \subset \mathbb{R}^n$ is such that $$\tilde{c}:I_\epsilon \to \varphi(U \cap V)$$ defined by $\tilde{c}(t) = \varphi \circ \psi^{-1} \circ c$:s image is in $\varphi(U \cap V)$.
Now, note that $\tilde{c}(0) = \varphi(p) = (x^1(p),\ldots,x^n(p)) \subset \partial \mathbb{R}^n\_$.
Lastly, we see, using the ordinary chain rule, that $$\tilde{c}(0)' = (\varphi \circ \psi \circ c)'(0) = D(\varphi \circ \psi^{-1})\Big|_{c(0)} \Big(c'(0)\Big) = D(\varphi \circ \psi^{-1})\Big|_{\psi(p)} (v) = e_1.$$ This means that the image of $$\tilde{c}'$$ is not in $\mathbb{R}^n\_$ for small $t \in I_{\epsilon}$. This leads to a contradiction.
Now, my question is this: What exactly is the contradiction here? In my notes, it´s says that the contradiction is that $\tilde{c}(t) \not \in \mathbb{R}^n\_$. Is the reason we can draw this conclusion, that $\tilde{c}'(0)$ is positive in the first argument? So this means that it is increasing in the first argument around $t = 0$? Since we concluded that $\tilde{c}(0) = p \in \partial \mathbb{R}^n\_$, this means in some small neighbourhood around $t = 0$, we have that $\tilde{c}(t) \in \mathbb{R}^n \setminus \mathbb{R}^n\_$? It is not explicit stated in the notes I have, but I believe an undisclosed assumption must be that $\varphi(U \cap V) \subset \mathbb{R}^n\_$? Because then this makes more sense.
Edit: There is however one thing I am unsure of, since we assume that $\varphi:U \to \varphi(U) \subset \mathbb{R}^n\_$ and $\psi:V \to \psi(V) \subset \mathbb{R}^n\_$, is not $$D(\varphi \circ \psi^{-1})|_{\psi(p)}: \mathbb{R}^n\_ \to \mathbb{R}^n\_?$$ Unless $\mathbb{R}^n\_$ is isomorphic to $\mathbb{R}^n$, I don´t see how $D(\varphi \circ \psi^{-1})_{\psi(p)}$ "hits" $e_1 \not \in \mathbb{R}^n\_$?
To sort of answer my own question, with some prompting from Paul Frost. I believe the context is that $M$ is a manifold with boundary, hence all charts $(U_i,\varphi_i)$ are such that $\varphi_i:U_i \to \varphi_i(U_i) \subset \mathbb{R}^n\_$. In the proof, we show that $y^1(p) < 0$ leads to a contradiction, hence we conclude that $y^1(p) = 0$ (the only alternative left). Hence we have shown what we set out to show.
Regarding my question why $$D(\varphi \circ \psi^{-1})\Big|_{\psi(p)}$$ "hits" $e_1$. I think that the explanation is that we have $$D(\varphi \circ \psi^{-1})\Big|_{\psi(p)}:\mathbb{R}^n \cong T_{\psi(p)}\psi(V) \to T_{\varphi(p)}\varphi(U) \cong \mathbb{R}^n.$$ That is, we take the codomain via an isomorphism to $\mathbb{R}^n$. I am not sure how to show this isomorphism though, at this point.