Consider the proof of Evans:
My questions:
Why the ball $B(x^{\epsilon})$ lies in $U\cap B(x^0,r)$, for all $x\in V?$ I think that this is true for an neighbourhood of $V$. How can i prove in a formal way and not in a visible way? Why Is importante the use of the ball? I think that the importante thing Is the point $x^{\epsilon}\in U$ for $x\in \epsilon-$ neighbourhood of $ V$. In this way i can answer to second question.
Why $v^{\epsilon}\in C^{\infty}(\overline{U})$?
$\newcommand{\eps}{\varepsilon}\def\ov{\overline}$Though I can't be sure, I suspect your problem lies in showing that $y_n > \gamma(y_1,\ldots,y_n)$ for $y \in B(x^\eps,\eps)$. This follows from $\gamma \in C^1$, or more precisely - from Lipschitz continuity $|\gamma(\ov{p})-\gamma(\ov{q})| \le L |\ov{p}-\ov{q}|$.
To explain this, let me use notation $\ov{p}=(p_1,\ldots,p_{n-1})$ for the first $n-1$ coordinates. Since $$ |\ov{y}-\ov{x}| = |\ov{y}-\ov{x^\eps}| \le |y-x^\eps| < \eps, $$ we have $$ \gamma(\ov{y}) \le \gamma(\ov{x}) + L \eps < x_n + L \eps. $$ On the other hand, $$ y_n > x^\eps_n - \eps = x_n + (\lambda-1)\eps. $$ It is easily seen that $y_n > \gamma(\ov{y})$ as long as $\lambda \ge L+1$.
With this, it should be clear that $B(x^{\eps},\eps)$ lies in $U \cap B(x^0,r)$ for every $x \in V$ (question (1)).
As for question (2), we have just observed that $u_\eps(x) := u(x^\eps)$ is well-defined not only for $x \in V$, but on a larger set. Precisely, on the $\eps$-neighborhood of $V$. Then it shouldn't be surprising that a convolution with $\eta \in C_c^\infty(B(0,\eps/2))$ is well-defined and smooth on $\ov{V}$ (or even the $\eps/2$-neighborhood of $V$).