Following the proof given in Milnor's Topology From A Differentiable Viewpoint:
$\require{AMScd}$
$\begin{CD}
x \in U \subseteq M @>F>> c \in V \subseteq N \\
@V\phi VV @VV\psi V \\
\phi\left(U\right) \subseteq H^{m} @>>\psi F \phi^{-1}> \psi\left(c\right) \in \psi\left(V\right)
\end{CD}$
Because $c \in N$ is a regular value of $F$, for every $x \in F^{-1}\left(c\right) \subseteq M$, there are charts $\left(U,\phi\right)$ at $x$ in $M$ and $\left(V,\psi\right)$ at $c$ in $N$ such that $\psi F \phi^{-1}: \phi\left(U\right) \subseteq H^{m} \to \psi\left(V\right) \subseteq \mathbb{R}^{n}$ is smooth, and has a regular value at $\psi\left(c\right)$.
$\begin{CD}
\phi\left(U\right) \subset W \subseteq \mathbb{R}^{m} @>G>> \psi\left(c\right) \in \mathbb{R}^{n}
\end{CD}$
Let $W$ be an open subset of $\mathbb{R}^{m}$ such that $W \cap H^{m} = \phi\left(U\right)$; and let $G:W\to\mathbb{R}^{n}$ be the smooth extension of $\psi F \phi^{-1}$ over $W$. Now, we can always choose $W$ small enough so that $G^{-1}\left(\psi\left(c\right)\right)$ does not contain any critical points (Sard's lemma; $\mathbb{R}^{m}$ is regular). Thus, $\psi\left(c\right)$ is a regular value of $G$; and by preimage theorem (for smooth manifolds), $Z := G^{-1}\left(\psi\left(c\right)\right)$ is a (m-n)-dimensional submanifold of $\mathbb{R}^{m}$.
Furthermore, since $G$ is constant over $Z$, $T_{a}Z \subseteq ker\left\{DG\left(a\right)\right\}$ for every $a \in Z$. But $DG\left(a\right):\mathbb{R}^{m}\to\mathbb{R}^{n}$ is surjective, and the rank-nullity theorem implies $dim \left(ker\left\{DG\left(a\right)\right\}\right) =$ (m-n). Thus, $T_{a}Z = ker\left\{DG\left(a\right)\right\}$.
Now, define $\pi: Z \subseteq \mathbb{R}^{m} \to \mathbb{R}$ as $\left(x_{1},\ldots,x_{m}\right) \mapsto x_{m}$.
To show that $0 \in \mathbb{R}$ is a regular value of $\pi$:
Suppose otherwise. That is, suppose $\exists$ $a \in Z \cap \partial H^{m}$ such that $d\pi_{a}:T_{a}Z \to \mathbb{R}$ is not surjective. Then, $ker \left\{d\pi_{a}\right\}$ $=$ $T_{a}Z$ $=$ $ker \left\{DG(a)\right\}$. But, we know that $ker \left\{d\pi_{a}\right\} \subseteq \mathbb{R}^{m-1} \times \left\{0\right\} = \partial H^{m}$. Thus, $ker \left\{DG(a)\right\} \subseteq \partial H^{m}$ if $0$ is not a regular value for $\pi$.
Now, since $c \in N$ is a regular value of $F|_{\partial M}$ as well, arguing as before, we can show that $\bar{G} := G|_{W\cap\partial H^{m}}$ has a regular value at $\psi\left(c\right)$. That is, for every $a \in Z\cap\partial H^{m}$, $D\bar{G}\left(a\right): \mathbb{R}^{m-1} \to \mathbb{R}^{n}$ is surjective; and by rank-nullity theorem, $dim \left(ker\left\{D\bar{G}\left(a\right)\right\}\right) = $ (m-n-1)
Finally, $ker \left\{DG(a)\right\} \subseteq \partial H^{m}$ implies $ker \left\{DG(a)\right\} = ker \left\{D\bar{G}(a)\right\}$, which is clearly false (dimension mismatch). Hence, $0$ must be a regular value for $\pi$.
Since $0 \in \mathbb{R}$ is a regular value for $\pi$, $\left\{z \in Z | \pi\left(z\right) \geq 0\right\}$ $=$ $\phi\left(U \cap F^{-1}\left(c\right)\right)$ is a manifold with boundary $\left\{z \in Z | \pi\left(z\right) = 0\right\}$ $=$ $\phi\left(U \cap F^{-1}\left(c\right) \cap \partial M \right)$.
$\phi$ being a diffeomorphism, $U \cap F^{-1}\left(c\right)$ is a manifold with boundary $U \cap F^{-1}\left(c\right) \cap \partial M$. Observing that this is true for every $x \in F^{-}\left(c\right)$ completes the proof.
Above we showed that the subset $\overset{\circ}{\mathcal E}_k$ of $\mathcal E_k$ whose elements have value in $[0,1)$ with respect the scalar function $g$ is open in $H^k_k$ since it is union of the sets
$$
g^{-1}\big[\,(0,1)\,\big]\cap H^k_k\,\,\,\text{and}\,\,\,C\Big(O,\frac 1k\Big)\cap H^k_k
$$
that are open in $H^k_k$. Now provided that
$$
\xi^1+\dots+\xi^k=1
$$
for any $\xi\in\mathcal E_k$ then surely it must be
$$
\xi^j\gneq0
$$
for any $j=1,\dots,k$ and in particular for now we assume that this happens for the last coordiante. So in this case we define the map $\varphi$ from $\Bbb R^k$ into $\Bbb R^k$ through the equation
$$
\varphi(x):=\big(x^1,\dots,x^{k-1},1-(x^1+\dots+x^{k-1}+x^k)\big)
$$
for any $x\in\Bbb R^k$ and thus we are going to prove that this map can be restricted to a coordinate patch about $\xi\in\mathcal E_k$ whose last coordinate is not zero, i.e. we let to prove that the restriction $\phi$ of $\varphi$ at the set $\overset{\circ}{\mathcal E}_k$ is the researched coordinate patch. First of all we observe that $\varphi$ is a diffeomorphism of $\Bbb R^k$ onto $\Bbb R^k$ being an affine map, i.e. it is compostion of a translation with a linear map between finite dimensional topological vector spaces and both these maps are diffeomorphism. So if we prove that $\varphi$ maps $\overset{\circ}{\mathcal E}_k$ into an open set of $\mathcal E_k$ containing $\xi$ then we will proved that $\varphi$ is a coordinate patch about $\xi$ so that we let to do this. Now if $x$ is an elemen of $\overset{\circ}{\mathcal F}$ then it must be
$$
x^i\in[0,1)
$$
for each $i=1,\dots,k$ since otherwise it would be
$$
x^1+\dots+x^{i-1}+1+x^{i+1}+\dots+x^k\le x^1+\dots+x^{i-1}+x^i+x^{i+1}+\dots+x^k<1\Rightarrow\\ x^1+\dots+x^{i-1}+x^{i+1}+\dots+x^k<0
$$
that is impossible if $x$ lies in $\overset{\circ}{\mathcal E_k}$ and thus in $H^k_k$ and so we can conclude that
$$
x\in\overset{\circ}{\mathcal E}_k\Rightarrow x\in[0,1)^k\wedge x^1+\dots+x^k\lneq1\Rightarrow\\\varphi(x)\in H^k_k\wedge\varphi^1(x)+\dots+\varphi^k(x)\le1\Rightarrow\varphi(x)\in\mathcal E_k
$$
and this proves that $\varphi$ maps $\overset{\circ}{\mathcal E}_k$ into $\mathcal E_k$. So if the last coordinate of $\xi$ is not zero and the others are not negative then surely
$$
g(\xi^1,\dots,\xi^{k-1},0)\ge0\,\,\,\text{and}\,\,\,g(\xi^1,\dots,\xi^{k-1},0)=\xi^1+\dots+\xi^{k-1}<\xi^1+\dots+\xi^{k-1}+\xi^k=1
$$
so that $(\xi^1,\dots,\xi^{k-1},0)$ is an element of $\overset{\circ}{\mathcal E}_k$ and in particular it is such that
$$
\varphi(\xi^1,\dots,\xi^{k-1},0)=\big(\xi^1,\dots,\xi^{k-1},1-(\xi^1+\dots+\xi^{k-1})\big)=\big(\xi^1,\dots,\xi^{k-1},\xi^k\big)=\xi
$$
and thus it is proved that $\xi$ is an element of $\varphi\big[\,\overset{\circ}{\mathcal E}_k\,\big]$. Now we previously define the set
$$
\tilde{\mathcal E}_k:=\{x\in\mathcal E_k:x^k\neq 0\}
$$
and thus we let prove that $\varphi$ carries $\overset{\circ}{\mathcal E}_k$ onto $\tilde{\mathcal E}_k$. So we first observe that
$$
x\in\overset{\circ}{\mathcal E}_k\Rightarrow 0\le g(x)\lneq1\Rightarrow0\lneq1-g(x)\le1\Rightarrow\varphi^k(x)>0\Rightarrow\varphi(x)\in\tilde{\mathcal E}_k
$$
and so we conclude that
$$
\varphi\big[\,\overset{\circ}{\mathcal E}_k\,\big]\subseteq\tilde{\mathcal E}_k
$$
so that we let to prove the opposite inclusion. Now for any element $\xi$ of $\tilde{\mathcal E}_k$ it must be
$$
g(\xi)=1\,\,\,\text{or either}\,\,\,g(\xi)<1
$$
but effectively we just proved above that if $g(\xi)=1$ and $\xi^k>0$ then surely
$$
\xi=\varphi(x)
$$
for any $x\in\overset{\circ}{\mathcal E}$ so that to follow we are supposing that $g(\xi)$ is strictly less than $1$. So we observe that
$$
\begin{cases}\xi\in\tilde{\mathcal E}_k\\
g(\xi)\lneq1\end{cases}\Rightarrow
\begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\\xi^k\gneq0\\
\xi^1+\dots+\xi^{k-1}+\xi^k\lneq1\end{cases}\Rightarrow\\
\begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\0\lneq\xi^k\lneq1\\
0\lneq1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\lneq1\end{cases}\Rightarrow
\begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\0\lneq1-\xi^k\lneq1\\
0\lneq1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\lneq1\end{cases}\Rightarrow
\begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\
1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\gneq0\\
0\lneq\xi^1+\dots+\xi^{k-1}+\big(1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\big)\lneq1\end{cases}
$$
and thus we conclude that $\big(\xi^1,\dots,\xi^{k-1},1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\big)$ is an element of $\overset{\circ}{\mathcal E}_k$ and so observing that
$$
\varphi\big(\xi^1,\dots,\xi^{k-1},1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\biggl)=\\
\biggl(\xi^1,\dots,\xi^{k-1},1-\Big(\xi^1+\dots+\xi^{k-1}+\big(1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\big)\Big)\biggl)=\\
\Biggl(\xi^1,\dots,\xi^{k-1},1-\Big((\xi^1+\dots+\xi^{k-1})+1-(\xi^1+\dots+\xi^{k-1})-\xi^k\Big)\Biggl)=\\
\big(\xi^1,\dots,\xi^{k-1},1-(1-\xi^k)\big)=(\xi^1,\dots,\xi^{k-1},\xi^k)=\xi$$
we conclude that $\varphi$ maps $\overset{\circ}{\mathcal E}_k$ onto $\tilde{\mathcal E}_k$. So we let to prove that $\tilde{\mathcal E}_k$ is open in $\mathcal E_k$ and we are doing this proving that the set $\mathcal E_k\setminus\tilde{\mathcal E}_k$ is closed in $\mathcal E_k$, that is it is the intersection of a closed set of $\Bbb R^k$ with $\mathcal E_k$. Now we remember that any symplex is closed (see here for details) so that the $(k-1)$ standard simplex $\mathcal E_{k-1}$ is closed: so observing that
$$
\mathcal E_k\setminus\tilde{\mathcal E}_k=\{x\in\mathcal E_k:x^k=0\}=\\
\{x\in\Bbb R^k:x^1+\dots x^{k-1}\le1\wedge x^i\ge0,\,\forall\,i=1,\dots,(k-1)\wedge x^k=0\}=\\
\{x\in\Bbb R^{k-1}:x^1+\dots+x^{k-1}\le1\wedge x^i\,\forall\,i=1,\dots,(k-1)\ge0\}\times\{0\}=\mathcal E_{k-1}\times\{0\}
$$
we conclude that $\mathcal E_k\setminus\tilde{\mathcal E}_k$ is closed in $\Bbb R^k$ (ideed it is product of closed set) and so $\mathcal E_k$ too. So we proved the existence of a coordinate patch when the last coordinate of $\xi$ is not zero. Otherwise if the last coordinate of $\xi$ is zero then surely (remember what observed above) there must exist $j=1,\dots,k-1$ such that
$$
\xi^j\neq0
$$
so that let be $\psi$ the diffeormosphism of $\Bbb R^k$ onto $\Bbb R^k$ that interchanges the $j$-th coordinate with the last, i.e.
$$
[\psi(x)](i):=\begin{cases}x^k,\,\,\,\text{if}\,\,\,i=j\\
x^j,\,\,\,\text{if}\,\,\,i=k\\
x^i\,\,\,\text{otherwise}\end{cases}
$$
for any $x\in\Bbb R^k$. Now $\psi$ is an involution that maps $\mathcal E_k$ onto $\mathcal E_k$ and in particular if the last coordinate of $\xi$ is zero then this does not happen for $\psi^{-1}(\xi)$ so that if $\varphi$ is a coordinate patch about $\psi^{-1}(\xi)$ then the composition $\psi\circ\varphi$ of $\psi$ with $\varphi$ is just a coordinate patch about $\xi$ and so the statement finally holds.
Best Answer
Here is a coordinate-free proof. Consider the inclusion $i : N \to M$. As $N$ is a submanifold, $i$ is an embedding, and for each $p \in N$, its differential $\mathrm{d}i_p : T_p N \to T_pM$ is injective, and identifies $T_pN$ as a linear subspace $\mathrm{d}i_p(T_pN) \subset T_pM$.