Assume $D \subset \mathbb{R}^{n}$ be a bounded and open domain with $C^{2}$ boundary. Let $x \in \partial D$ and $r>0$. Define \begin{align*} C_{r} := D \cap \partial B( x,r) \text{ and } \overline{C_{r}}:= \{y \in \partial B( x,r ): y \cdot \nu( y )<0 \} .\end{align*} WLOG, we may ssume that $x=0$ and the normal derivative $n( x ) $ at $x$ is given by $( 0, \ldots ,0 ,-1 ) $. The goal is to show that \begin{align*} \int_{C_{r}} d S_{y} = \int_{\overline{C_r}} dS_{y} + O( r^{n} ) .\end{align*}
This is a proof related to Gauss lemma. https://web.stanford.edu/class/math220b/handouts/potential.pdf (page 8, Claim 3).
I think the idea is when $r$ approaches $0$, $\overline{C_r}$ approaches the half circle of $\partial B(x,r)$. I am a bit lost in first paragraph of the proof.
Using the notation from: https://web.stanford.edu/class/math220b/handouts/potential.pdf
They are arguing that surface area of $C_{\epsilon} \setminus \tilde{C_{\epsilon}}$ is bounded by surface area of a strip on the surface of $B(x,\epsilon)$ and hence can be bounded by area of a rectangle along the surface of $B(x,\epsilon)$ where base length is the circumference of the $B(x,\epsilon)$ and hence is the volume of a $n-2$ dimensional sphere of radius $\epsilon$ and hence $O(\epsilon^{n-2})$. Now height of the strip or rectangle is obtained as follows. The boundary $\Omega$ is a $C^2$ boundary and hence is defined by $\partial \Omega \cap B(x,\epsilon) = \{(y_1,...,y_{n-1},y_n) : y_n = f(y_1,...,y_{n-1})\}$. This is by definition.
Now assume $x = 0$ is the origin and hence the height is the value of $y_n$ at distance $\epsilon$ from origin (on the surface of $B(x,\epsilon)$) and $y_n$ is bounded by $f(y_1,...,y_{n-1})$ evaluated with $(y_1,...,y_{n-1})$ on the $n-1$ dimensional sphere of $B(x,\epsilon)$ which is passing through $x$ and lies in $v(x).y =0$ i.e., $y_n =0$. Since $y_n > 0$ (by the fact $y.v(x) < 0$), we have $|y_n| \leq |f(y_1,...,y_{n-1})|$. Now apply Taylor series to $n-1$ variable function $f$ and write $$f(y_1,...,y_{n-1}) = f(0)+\nabla f(0) (y_1,...,y_{n-1})^T$$ $$+(y_1,...,y_{n-1}) Hessian(f(\gamma)) (y_1,...,y_{n-1})^T$$. Since $Hessian(f(\gamma))$ is bounded on a compact set, we have $$|f(y_1,...,y_{n-1})| \leq |f(0)| + |\nabla f(0)| |(y_1,...,y_{n-1})| $$ $$+ \sup_{y_1,..,y_{n-1}: |(y_1,...,y_{n-1})| \leq \epsilon} |Hessian f(y)| \times |(y_1,..y_{n-1})|^2 \leq C \epsilon^2$$
This is because they assumed $f(0) = \nabla f(0) = 0$ and $Hessian(f)|_{y_1,y_2,...,y_{n-1}=0}$ is full rank.