Surface measure $\mathrm{d}S$ of $M$ corresponding to parametrisation I'm reading these notes, and there's a part I don't understand on page 72 (76 in the PDF), right after the equation marked with two stars.

Note that $\sqrt{1+|Dg(y')|^2}dy'$ is the surface measure $\mathrm{d}S$ of $M$ corresponding to the parametrisation $y'\mapsto(g(y'),y')$ of $M$

A reference is then given, but I can only find it in German and can't understand it.
$y'$ is some point in $\mathbb{R}^{n-1}$, $M$ is a hypersurface in an open subset of $\mathbb{R}^n$. $g$ is a function from $\mathbb{R}^{n-1}$ to $\mathbb{R}$.

This hypersurface $S$ is parametrized by $$y'\mapsto (y',g(y'))\in{\mathbb R}^n\qquad(y'\in{\mathbb R}^{n-1})\ ,$$
hence is considered as graph of the scalar function $g$ defined on ${\mathbb R}^{n-1}$.
If $S$ were a hyperplane $y_n=a_1y_1+a_2y_2+\ldots a_{n-1}y_{n-1}$ then the surface measure on $S$ would be ${1\over\cos\phi}{\rm d}(y')$, where ${\rm d}(y')$ is the $(n-1)$-dimensional euclidean (or Lebesgue) measure on the coordinate hyperplane $y_n=0$, and $\phi$ is the angle between the normal of $S$ and the $y_n$-axis. Now
$${1\over\cos\phi}={1\cdot\sqrt{1+a_1^2+a_2^2+\ldots+a_{n-1}^2}\over\langle{\bf e}_n\cdot(-a_1,-a_2,\ldots,-a_{n-1},1)\rangle}=\sqrt{1+a_1^2+a_2^2+\ldots+a_{n-1}^2}=\sqrt{1+|{\bf a}|^2}\ .$$
In the case at hand the tangent plane $T_p$ to the curved surface $S$ at some point $p\in S$ is given by
$$T_p:\quad y_n-p_n=\nabla g(p')\cdot (y'-p')\ .$$
The local factor ${1\over\cos\phi}$ therefore is given by $\sqrt{1+|\nabla g(y')|^2}$, as indicated in your source.
