The following is theorem 4.15 of Lee's Introduction to smooth manifolds
Theorem 4.15. Suppose $M$ is a smooth $m$-manifolds with boundary, $N$ is a smooth $n$-manifold, and $F:M\to N$ is a smooth immersion. For any $p\in\partial M$, there exists a smooth boundary chart $(U,\varphi)$ for $M$ centered at $p$ and a smooth coordinate chart $(V,\psi)$ for $N$ centered at $F(p)$ with $F(U)\subset V$, in which $F$ has the coordinate representation
$$\hat{F}(x^1,…,x^m) = (x^1,…,x^m,0,…,0).\ (\dagger)$$
Proof. By choosing initial smooth charts for $M$ and $N$, we may assume that $M$ and $N$ are open subsets of $\Bbb H^m$ and $\Bbb R^n$, respectively, and also that $p = 0\in\Bbb H^m$, and $F(p) = 0\in\Bbb R^n$. By definition of smoothness for functions on $\Bbb H^m$, $F$ extends to a smooth map $\tilde{F}:W\to\Bbb R^n$, where $W$ is an open subsets of $\Bbb R^m$ containing $0$. Because $d\tilde{F}_0 = dF_0$ is injective, by shrinking $W$ if necessary, we may assume that $\tilde{F}$ is a smooth immersion. By rank theorem, there exist smooth charts $(U_0,\varphi_0)$ for $\Bbb R^m$ centered at $0$ and $(V_0,\psi_0)$ for $\Bbb R^n$ centered at $0$ such that $\hat{F} = \psi_0\circ\tilde{F}\circ\varphi_0^{-1}$ is given by $(\dagger)$. The only problem with these coordinates is that $\varphi_0$ might not restrict to a boundary chart for $M$…
Here, what is the meaning of $\varphi_0$ might not restrict to a boundary chart for $M$?
Best Answer
The definition of "boundary chart" is on page 25: it's a homeomorphism taking an open subset of $M$ to an open subset of the upper half-space $\mathbb H^m$ that has nonempty intersection with the boundary of $\mathbb H^m$ (and therefore takes boundary points of $M$ to $\partial \mathbb H^m$).
The question is whether the restriction of $\varphi_0$ to $M$ (which we are assuming to be an open subset of the upper half-space $\mathbb H^n$) is a boundary chart, meaning it would have to take its values in the upper half space $\mathbb H^m$, sending boundary points of $M$ to the boundary of $\mathbb H^m$. Since $\varphi_0$ was obtained by applying the rank theorem on an open subset of $\mathbb R^m$, there's no reason why $\varphi_0$ should satsify this property.