Flattening a hypersurface Let $\overline{M}^{n+1}$ be a Riemannian manifold with metric $g$ and $M^n$ be an embedded orientable hypersurface of $\overline{M}$ with unitary normal $\eta : M \to T\overline{M}$. 
For a fixed point $p$ of $M$ I know there exists a local chart $\varphi : U \ni p \to V \subset \mathbb{R}^{n+1}$ such that


*

*$\varphi(p) = 0$

*$\varphi(U \cap M) = V \cap (\mathbb{R}^n \times \{0\})$


My question is: can we further assume $d \varphi_p (\eta(p)) = e_{n+1} = (0, \dots, 0, 1)$?
 A: Yes we can.
Here is a way to do it, it is actually quite easy. Assume you have a local chart $\varphi$ as you wrote, which does not necessarily satisfy what you want. Let us fix it, or rather let us fix its inverse $\psi := \varphi^{-1}$, like so: define $\psi_1: V \to \overline{M}$ by
$$\psi_1(x_1, \dots, x_n, t) = \exp_{m}(t \, \eta(m))$$
where $m = \psi(x_1, \dots, x_n, 0)$.
This map $\psi_1$ is well-defined after maybe making $V$ smaller in the last coordinate, because the Riemannian exponential map is well-defined in a neighborhood of $0$. It is immediate to check that $\psi_1$ has the following properties:


*

*$\psi_1 = \psi$ on $V \cap \mathbb{R}^n$.

*$\psi_1$ is differentiable, and its differential at $0$  is given by $d\psi_1 (e_k) =  d \psi(e_k)$ for $1 \leqslant k \leqslant n$, and $d\psi_1 (e_{n+1}) =  \eta (p)$.


In particular, the differential of $\psi_1$ at $0$ is bijective because it has full rank: 
\begin{align*}
d\psi_1(\mathbb{R}^{n+1}) &= d\psi_1(\mathbb{R}^{n}) + d\psi_1(\mathbb{R}e_{n+1}) \\
&= d\psi(\mathbb{R}^{n}) + \mathbb{R}\,d\psi_1(e_{n+1}) \\
&= T_pM + \mathbb{R}\,\eta(p) \\
&= T_p \overline{M}~.
\end{align*}
By the inverse function theorem, after making $V$ even smaller, we can assume that $\psi_1$ is a diffeomorphism. Finally, take the chart $\varphi_1 := {\psi_1}^{-1}$.
Remark: Note that I used the Riemannian exponential map to construct $\psi_1$ for convenience, but I could instead have used the flow of any vector field which extends $\eta$. In fact, your question does not have much to do with Riemannian geometry, it does not matter that $\eta$ is normal to $M$, any transverse vector field would do just as fine.
