The key concept from Differential Topology is that of a regular domain, defined for instance in Ch. 5 of John Lee's Smooth Manifolds 2nd Edition. In the smooth setting, the equivalence is described Theorem 5.48 etc. While these proofs are in the setting of Smooth Manifolds, it should all go through exactly the same in the setting of $C^1$ manifolds. Below is an attempt at a more direct proof using the implicit function theorem.
Definition 1 $\implies$ Definition 2 is relatively straightforward. We abbreviate $(x_1,\dots,x_N)=(x',x_N)$. Then, define $\phi(x',x_N)=(x',x_N-\gamma(x'))$. This will be a $C^1$-diffeomorphism as the inverse is $(x',x_N)\mapsto (x',x_N+\gamma(x'))$.
For Definition 2 $\implies$ Definition 1, suppose we are given $\varphi(x)$ and write the components as $\varphi=(\varphi_1,\dots,\varphi_N)$. As $\varphi$ is a diffeomorphism, we must have that $\vec{\nabla} \varphi_N\mid_{x_0}\neq \vec{0}$. Thus, it must have some non-zero component, $\frac{\partial \varphi_N}{\partial x_j}\neq 0$. By reordering and reorienting the axes, we may assume $\frac{\partial \varphi_N}{\partial x_N}\mid_{x_0}> 0$. By shrinking the neighboorhood about $x$, we may assume $\frac{\partial \phi_N}{\partial x_N}>0$ on all of $B(x_0,r)$. By the implicit function theorem applied to the map $\phi_N(x',y)$ (for instance as stated in the appendix of Evans), there is a function $\gamma$ such that
$$
(x',x_N)\in \partial U\iff \Phi_N(x',x_N)=0\iff x_N-\gamma(x')=0.
$$
We are done if we can show that $(x',x_N)\in U\iff x_N>\gamma(x')$ (this step may be where people are struggling). This is true by our reduction to the case of $\frac{\partial \varphi_N}{\partial x_N}\mid_{x_0}> 0$. As $\varphi_N(x',\cdot)$ is increasing on $B(x_0,r)$, we have that
$$
\varphi_N(x',x_N)>0\iff \varphi(x',x_N)>\varphi(x',\gamma(x'))\iff x_N>\gamma(x').
$$
You have a lot of freedom when you choose cut-off functions, so much so you can usually impose conditions on their gradients. Here's what I mean: let $\zeta \in C^\infty(\mathbb{R})$ be such that $$
\begin{cases}
\zeta(t) = 1 \text{ if } t \leqslant 1, \, \zeta(t) = 0 \text{ if } t \geqslant 2 \\
0 \leqslant \zeta \leqslant 1 \\
0\leqslant \zeta' \leqslant 2
\end{cases}
$$ and define for $k\in \mathbb{N}$, $\zeta_k(x) = \zeta (\vert x \vert /k)$. Then $\zeta_k$ satisfy
$$
\begin{cases}
\zeta_k(t) = 1 \text{ in } B_k, \, \zeta_k(t) = 0 \text{ in } \mathbb{R}^n \backslash B_{k+1} \\
0 \leqslant \zeta_k \leqslant 1 \\
0\leqslant \vert D \zeta_k \vert \leqslant 2/k .
\end{cases}$$
Since $\zeta_k u $ satisfy the assumptions of the previous lemma in Brezis, it follows $\zeta_k u \in W^{1,p}_0(\Omega)$. Then since $\vert\zeta_k u - u \vert^p \leqslant \vert u \vert^p$, it follows from dominated convergence that
\begin{align*}
\| \zeta_k u - u \|_{L^{p}(\Omega)}^p &= \int_\Omega \vert\zeta_k u - u \vert^p d x \to 0 \qquad \text{ as } k \to \infty.
\end{align*} Moreover, \begin{align*}
\| \nabla(\zeta_k u) - \nabla u \|_{L^{p}(\Omega)} &= \bigg ( \int_\Omega \vert \nabla(\zeta_k u) - \nabla u\vert^p d x \bigg) ^{1/p} \\
&\leqslant \bigg ( \int_\Omega \vert u\nabla\zeta_k \vert^p d x \bigg) ^{1/p} + \bigg ( \int_\Omega \vert \zeta_k \nabla u - \nabla u\vert^p d x \bigg) ^{1/p} \\
&\leqslant\frac 2 k \bigg ( \int_\Omega \vert u \vert^p d x \bigg) ^{1/p} + \bigg ( \int_\Omega \vert \zeta_k \nabla u - \nabla u\vert^p d x \bigg) ^{1/p}\\
&\to 0 \qquad \text{ as } k \to \infty.
\end{align*} (Dominated convergence was used again for the second integral on the right hand side). Since $W^{1,p}_0(\Omega)$ is closed in $W^{1,p}(\Omega)$, it follows $u \in W^{1,p}_0(\Omega)$.
Best Answer
The definitions are equivalent. Brezis says: locally, the domain can be mapped onto half-space by a $C^1$-diffeomorphism. Evans says: locally, the domain is bounded by the graph of a $C^1$-function.
Evans$\implies$Brezis: the domain $x_n>\gamma(x_1,\dots,x_{n-1})$ is mapped onto upper halfspace by $$H(x_1,\dots,x_n) = x_1,\dots,x_{n-1}, x_n - \gamma(x_1,\dots,x_{n-1})$$ This is indeed a diffeomorphism, because $$H^{-1}(x_1,\dots,x_n) = x_1,\dots,x_{n-1}, x_n + \gamma(x_1,\dots,x_{n-1})$$
Brezis$\implies$Evans: by the implicit function theorem, the set where then $n$th component of $H$ is $0$ is a smooth hypersurface. Make a tangent plane to it a coordinate hyperplane; you get (again from IFT) a smooth function $\gamma$ for which this surface is a graph.