So clearly $M$ without corner points is equal to the set $\overset{\circ}M\cup\partial_1M$ so that we let to prove the statement finding a coordinate patch for the elements of $\overset{\circ}M$ and for the elements of $\partial_1M$ showing that the formers are interior points and the latters are boundary points.
Now we let to prove that it is possible to restrict any coordinate patch about $y\in\overset{\circ}M$ in an open neighborhood in $M$ contained in $\overset{\circ}M$. So if $\alpha:U\rightarrow V$ is a smooth coordinate patch about $y\in\overset{\circ}M$ then the statement is surely true if $U$ is open in $\Bbb R^k$ so that we are supposing that it is open in $H^k$ but not in $\Bbb R^k$. So if $U$ is open in $H^k$ then it is equal to the intersection of an open set $A$ of $\Bbb R^k$ with $H^k$ so that the set
$$
U^+:=U\cap\operatorname{int}H^k=(A\cap H^k)\cap\operatorname{int}H^k=A\cap\big(\,H^k\cap\operatorname{int}H^k\big)=A\cap\operatorname{int}H^k
$$
is a not empty (actually $U^+$ is the interior of $U$ in $\Bbb R^k$ and it is surely not empty because $H^k$ is convex and the interior of a not empty open set in a convex set is not empty as here it is showed) open set of $\Bbb R^k$ contained in $U$ and thus its immage $V^+$ under $\alpha$ is open in $M$ and contained in $\overset{\circ}M$. So we let to prove that any open set $W$ of $M$ contained in $\overset{\circ}M$ is an open set of $\overset{\circ}M\cup\partial_1M$ too. So first of all we observe that the sets $\overset{\circ}M$ and $\partial M$ are disjoint (indeed otherwise there would exist two coordinate patch $\alpha_i:U_i\rightarrow V_i$ about $y\in M\cap\partial M$ that would be respectively defined in an open set $U_1$ of $\Bbb R^k$ and in an open set $U_2$ of $H^k$ but not in $\Bbb R^k$ so that by the inverse function theorem an open set of $\Bbb R^k$ would be diffeomorphic to an open set of $H^k$ but not in $\Bbb R^k$ -in fact for the mentioned theorem the transition function $\alpha^{-1}_2\cap\alpha_1$ is a local diffeomorphism- and this is impossible) so that if $A$ is an open set of $\Bbb R^n$ whose intersection with $M$ is equal to $W$ then
$$
A\cap\partial M=\emptyset
$$
because
$$
A\cap\partial M=(A\cap\partial M)\setminus\overset{\circ}M=\big((A\cap\overset{\circ} M)\setminus\overset{\circ}M\big)\cup\big((A\cap\partial M)\setminus\overset{\circ}M\big)=\\
\big((A\cap\overset{\circ}M)\cup(A\cap\partial M)\big)\setminus\overset{\circ}M=\big(A\cap(\overset{\circ}M\cup\partial M)\big)\setminus\overset{\circ}M=(A\cap M)\setminus\overset{\circ}M=W\setminus\overset{\circ}M
$$
and $W$ is contained in $\overset{\circ}M$. Now $\partial_1 M$ is obviously contained in $\partial M$ so that $A$ and $\partial_1M$ are disjoint and thus
$$
A\cap(\overset{\circ}M\cup\partial_1M)=(A\cap\overset{\circ}M)\cup(A\cap\partial_1M)=(A\cap\overset{\circ}M)\cup\emptyset=(A\cap\overset{\circ}M)\cup(A\cap\partial M)=\\
\big(A\cap(\overset{\circ}M\cup\partial M)\big)=A\cap M=W
$$
which implies that $W$ is an open set of $(\overset{\circ}M\cup\partial_1M)$. So we conclude that the restriction $\alpha^+$ of $\alpha$ to $U^+$ is a smooth coordinate patch for $\overset{\circ}M\cup\partial_1M$.
Now any element $\eta$ of $\partial_1M$ has a smooth coordinate patch $\alpha:U\rightarrow V$ defined in a set $U$ that is open in $H^k$ but not in $\Bbb R^k$ and in particular
$$
\eta=\alpha(\xi)
$$
for any $\xi\in U$ such that
$$
\xi(j)=0\Leftrightarrow j=i
$$
for only one $i=1,\dots,k$. So for this such $\xi$ let be
$$
\mu:=\max\{\xi_i:i=1\dots,k\}>0
$$
so that
$$
\xi\in U_0:=\big(\,(0,\epsilon)^{i-1}\times[0,\epsilon)\times(0,\epsilon)^{k-i}\,\big)\cap U
$$
for any $\epsilon>\mu$. Now we observing that
$$
U_0:=\big(\,(0,\epsilon)^{i-1}\times[0,\epsilon)\times(0,\epsilon)^{k-i}\,\big)\cap U=\Big(\big(\,(0,\epsilon)^{i-1}\times(-\epsilon,\epsilon)\times(0,\epsilon)^{k-i}\,\big)\cap H^k\Big)\cap U
$$
and thus we conclude that $U_0$ is a subset of $U$ that is open in $U$ and in $H^k$ because $U$ is an open set of $H^k$ and so the subspace topology on $U$ of $\Bbb R^k$ is the same of $H^k$ so that the statement follows directly observing that $U_0$ is the intersection between an open set of $H^k$ with $U$ just like we proved in the above identity; moreover we observe that $U_0$ is contained in $(\Bbb R^+)^{i-1}\times\Bbb R^+_0\times(\Bbb R^+)^{k-i}$ and it is there open too because $U_0$ and $(\Bbb R^+)^{i-1}\times\Bbb R^+_0\times(\Bbb R^+)^{k-i}$ are open in $H^k$ so that their intersection is open in $H^k$ and so in $(\Bbb R^+)^{i-1}\times\Bbb R^+_0\times(\Bbb R^+)^{k-i}$ due to analogous arguments above advanced. Therefore the immage $V_0$ of $U_0$ under $\alpha$ is open in $M$ and moreover this application maps $U_0$ into $\overset{\circ}M\cup\partial_1M$. Now we let to prove that any open set $W_0$ of $M$ contained in $\overset{\circ}M\cup\partial_1 M$ is there open too. So first of all we observe that the sets $\partial M\setminus\partial_1M$ and $\overset{\circ}M\cup\partial_1 M$ are disjoint (indeed $\partial M\setminus\partial_1M$ is contained in $\partial M$ that is disjoint from $\overset{\circ} M$ and obviously it is disjoint from $\partial_1M$ too) so that if $A_0$ is an open set of $\Bbb R^n$ whose intersection with $M$ is equal to $W_0$ then
$$
A_0\cap(\partial M\setminus\partial_1M)=\emptyset
$$
because
$$
A_0\cap(\partial M\setminus\partial_1M)=\big(A_0\cap(\partial M\setminus\partial_1M)\big)\setminus(\overset{\circ}M\cup\partial_1M)=\\
\Big(\big(A_0\cap(\overset{\circ}M\cup\partial_1M)\big)\setminus(\overset{\circ}M\cup\partial_1M)\Big)\cup\Big(\big(A_0\cap(\partial M\setminus\partial_1M)\big)\setminus(\overset{\circ}M\cup\partial_1M)\Big)=\\
\Big(\big(A_0\cap(\overset{\circ}M\cup\partial_1M)\big)\cup\big(A_0\cap(\partial M\setminus\partial_1M)\big)\Big)\setminus(\overset{\circ}M\cup\partial_1M)=\\
\Big(A_0\cap\big((\overset{\circ}M\cup\partial_1 M)\cup(\partial M\setminus\partial_1M)\big)\Big)\setminus(\overset{\circ}M\cup\partial_1M)=\\
(A_0\cap M)\setminus(\overset{\circ}M\cup\partial_1M)=W_0\setminus(\overset{\circ}M\cup\partial_1M)
$$
and $W_0$ is contained in $(\overset{\circ}M\cup\partial_1M)$. So we observe that
$$
A_0\cap(\overset{\circ}M\cup\partial_1M)=\big(A_0\cap(\overset{\circ}M\cup\partial_1M)\big)\cup\emptyset=
\big(A_0\cap(\overset{\circ}M\cup\partial_1M)\big)\cup\big(A_0\cap(\partial M\setminus\partial_1M)\big)=\\
\Big(A_0\cap\big((\overset{\circ}M\cup\partial_1M)\cup(\partial M\setminus\partial_1M)\big)\Big)=A_0\cap\Big(\overset{\circ}M\cup\big(\partial_1M\cup(\partial M\setminus\partial_1M)\big)\Big)=\\
A_0\cap(\overset{\circ}M\cup\partial M)=A_0\cap M=W_0
$$
and thus the statement follows. So finaly we rememeber that the sets $(\Bbb R^+)^{i-1}\times\Bbb R^+_0\times(\Bbb R^+)^{k-i}$ and $\Bbb R^{k-1}\times\Bbb R^+_0$ are diffeomorphic (indeed the sets $\Bbb R^+$ and $\Bbb R$ are diffeomorphic via the logaritmic and exponential functions respectively) and thus if $\phi_i$ is a diffeomorphism between the latter and the former then the restriction of $\alpha\circ\phi_i$ to $\phi_i^{-1}[U_0]$ is a smooth coordinate patch of $\overset{\circ}M\cup\partial_1M$ defined in an open set of $\Bbb R^{k-1}\times\Bbb R^+_0$.
