I don't understand the proof that $ \| \overline{u} \|_{W^{1,p}(B)} \leq C \| u \|_{W^{1,p}(B^+)}$ This is from Evan's PDE's Extension Theorem (Theorem 1 in Section 5.4.) One can find a proof like this here. I have an old edition of the book so I don't think my page is going to work for anyone here.
Here are the relevant information
Define
$$\overline{u}(x) = \left\{\begin{matrix}
u(x) & x \in B^+ \\ 
 -3u(x_1,\dots,x_{n-1},-x_n) + 4u(x_1,\dots,x_{n-1},-x_n/2)& x \in B^-
\end{matrix}\right.$$
where $B$ is an open ball with centre $x^0$ and radius $r$ and
$$\left\{\begin{matrix}
B^+ := B \cap \{x_n \geq 0\}  \subset \overline{U}
\\ 
B^- := B \cap \{x_n \leq 0\} \subset R^n -\overline{U}
\end{matrix}\right.$$
where $U$ is a bounded (open) set.
$W^{1,p}(U)$ is the Sobolev space of $U$.
Q1: In the pdf I linked (Step 4), why is $\| \overline{u} \|_{W^{1,p} (B) } =\|-3u(...,-x_n) + 4u(...,-x_n/2) \|_{W^{1,p} (B) }$? Isn't the quantity $-3u(...,-x_n) + 4u(...,-x_n/2)$ only defined on the lower half of the ball?



Q2: What exactly is $\| \overline{u} \|_{W^{1,p} (B) }?$ Is it defined as
$$\int_B \| \overline{u} \|^p + \int_B \|D\overline{u} \|$$?
And is that further equal to
$$\int_B \| \overline{u} \|^p = \int_{x_n = 0}  \| \overline{u} \|^p + \int_{B^{+}} \| u \|^p +  \int_{B^{-}} \| -3u + 4u \|^p$$

and similarly for $\int_B \|D\overline{u} \|.$
Q3: Lastly I just don't quite understand the last estimation involving $B \cap \{x_n < r\}$
 A: re:Q2, Evans defines $$\|f\|_{W^{1,p}(B)}^p :=\sum_{0\le\alpha\le 1}\|D^\alpha f\|_{L^p(B)}^p= \sum_{0\le\alpha\le 1}\int |D^\alpha f(x)|^p \ dx.$$ Note $\|f\|_X$ is usually reserved for a function norm; the Euclidean norm / absolute value is written $|x|$.
Note that it is different from
\begin{align} \| f\|_{L^p(B)}^p+\|D f\|_{L^p(B)}^p
&= \int_B | f(x)|^p \ dx + \int_B | Df(x)|^p \ dx \\
&= \int_B | f(x)|^p \ dx +\int_B \left(\sum_{|\alpha|=1}| D^\alpha f(x)|^2\right)^{p/2} \ dx ,\end{align}
but they define equivalent norms. What you wrote with $\int |Du| dx$ is NOT equivalent because its missing a $p$. Unless $p=1$.
So re:Q1, you have indeed found a typo; $u$ is only defined on the upper ball. In fact there are more typos, the notes write $\alpha<1$ but this means $\alpha=0$. And the calculation involves $|2^{-p}v|^p$ which is $2^{-p^2}|v|^p$? It may be better to read Evans (sorry Chee Han!). Perhaps what they meant to write was
$$ \|\bar u\|_{W^{1,p}(B^-)} = \|{-3} u(\dots,-x_n)+4 u(\dots,-x_n/2)\|_{W^{1,p}(B^-)}.$$
In order to use that $\|\bar u\|_{W^{1,p}(B^-)} + \|\bar u\|_{W^{1,p}(B^+)} =  \|\bar u\|_{W^{1,p}(B)} $, for which we first need weak differentiability, one uses some continuity arguments (which you have because of the way the function was extended to the lower ball; this is the earlier part of the proof in Evans.)
Back to Q2, specialising to $f=\overline u$:
$$\int_B | \overline{u} |^pdx =  \int_{B^{+}} | u |^p dx+  \int_{B^{-}} | {-3}u(x',-x_n) + 4u(x',-x_n/2) |^p dx' dx_n.$$
We used that $B=B^+\cup B^-$ up to a $\mathbb R^n$-Lebesgue null set (a subset of $\{x:x_n=0\}$).
Note that the last term $\int_{B^{-}} | {-3}u(x',-x_n) + 4u(x',-x_n/2) |^p dx' dx_n$ is different from
$$ \int_{B^{-}} | {-3}u + 4u |^p dx= \int_{B^{-}} | {-3}u(x',x_n) + 4u(x',x_n) |^p dx' dx_n,$$
which is what you wrote. As you noted, $u$ is not defined for $x_n<0$, so this doesn't make sense. It is useful to write things out fully, instead of trying to use short notation.
As for Q3, using $$ |x+y|^p \le 2^{p-1}( |x|^p + |y|^p),$$
(this follows by the convexity of $t\mapsto t^p$ on $[0,\infty)$) you can split $\int_{B^{-}} | {-3}u(x',-x_n) + 4u(x',-x_n/2) |^p dx' dx_n$ into two types of integrals, times a constant depending on $p$:
One is
$$I_1=\int_{B^{-}} | u(x',-x_n)  |^p dx' dx_n = \int_{B^{+}} | u(x',y_n)  |^p dx' dy_n, $$
(by the obvious change of variables $B^- \to B^+$) and the other is
$$
I_2=\int_{B^{-}} | u(x',-x_n/2)  |^p dx' dx_n = 2\int_{V} | u(x',z_n)  |^p dx' dz_n ,$$
where $z_n = -x_n/2, dx_n = 2dz_n$ (as unsigned measures), and $$V:=\{ (x_1,\dots,x_{n-1},x_n/2) : x\in B^+\}.$$
Note that $V\subset B^+$; this means we can bound $I_2$ by
$$ I_2 \le 2\int_{B^+} | u(x)  |^p dx.$$
In total you should get for the $L^p$ norm,
$\| \overline u\|_{L^p(B)} \le C_p \|u\|_{L^p(B^+)}$ with $C_p=1+2^{p-1}(3+4\times 2).$ But what's important is that $C_p < \infty$, not the exact value.
This set $V$ is the correct one to use; note carefully it is different from $B^+\cap \{ x_n < r/2\}$ as in the screenshot.
For the derivatives, its the same calculation. The only thing is to be careful with chain rule if you want to compute the constant correctly: for $x\in B^-$,
$$ \partial_{x_n} \overline u(x) =  \partial_{x_n}(-3u(x',-x_n)+4u(x',-x_n/2)) = 3 (\partial_{x_n} u)(x',-x_n)-2 (\partial_{x_n} u)(x',-x_n/2)$$
The amusing fact that  $-3+4=1=3-2$, which you should have already seen when verifying the statements in the earlier part of the proof, is what makes the particular choice of "reflection" work.
A: *

*Yes. For example, $u(x_{1},\ldots,x_{n-1},-x_{n})$ is not defined if $x_{n}>0$.

*

$$\|\overline{u}\|_{W^{1,p}(B)}:=\left( \sum_{|\alpha| \leq 1} \int_{B}|D^{\alpha}\overline{u}|^{p}dx\right)^{\frac{1}{p}},$$
where $\alpha=(\alpha_{1},\ldots,\alpha_{n}) \in \mathbb{Z}_{\geq 0}^{n}$ is a multiindex and $|\alpha|=\sum_{i=1}^{n}\alpha_{i}$ is its norm. If you expand this, you obtain
$$\|\overline{u}\|_{W^{1,p}(B)}=\left( \sum_{|\alpha| \leq 1} \int_{B}|D^{\alpha}\overline{u}|^{p}dx\right)^{\frac{1}{p}}=\left(\int_{B}|\overline{u}|^{p}dx+\sum_{i=1}^{n}\int_{B}|\overline{u}_{x_{i}}|^{p}dx\right)^{\frac{1}{p}}.$$
Also
$$\int_{B}|\overline{u}|^{p}dx=\int_{B^{+}}|u|^{p}dx+\int_{B^{-}}|-3u(x_{1},\ldots,x_{n-1},-x_{n})+4u(x_{1},\ldots,x_{n-1},-x_{n}/2)|^{p}dx. $$
The integral over $\{x_{n}=0\}$ vanishes because this set has Lebesgue measure zero.


*Their stimations do not have sense because $u(\ldots,-x_{n}/2)$ is only defined for $B^{+}$, not for the whole ball.

Finally, let me prove that $\|\overline{u}\|_{W^{1,p}(B)} \leq C \|u\|_{W^{1,p}(B^{+})}$.
Note that
\begin{align*}
\int_{B}|\overline{u}|^{p}dx &=\int_{B^{+}}|u|^{p}dx+\int_{B^{-}}|-3u(x_{1},\ldots,x_{n-1},-x_{n})+4u(x_{1},\ldots,x_{n-1},-x_{n}/2)|^{p}dx \\
&\leq \int_{B^{+}}|u|^{p}dx+3\int_{B^{-}}|u(x_{1},\ldots,x_{n-1},-x_{n})|^{p}dx+4\int_{B^{-}}u(x_{1},\ldots,x_{n-1},-x_{n}/2)|^{p}dx \\
&=\int_{B^{+}}|u|^{p}dx+3\int_{B^{+}}|u(y_{1},\ldots,y_{n-1},y_{n})|^{p}dy+2\int_{B^{+}}u(y_{1},\ldots,y_{n-1},y_{n})|^{p}dy, \\
&=6\int_{B^{+}}|u|^{p}dx.
\end{align*}
where in the penultimate inequality I used the substitutions $(y_{1},\ldots,y_{n})=(x_{1},\ldots,-x_{n})$ and $(y_{1},\ldots,y_{n})=(x_{1},\ldots,-x_{n}/2)$, respectively. Also,
\begin{align*}
\sum_{i=1}^{n}\int_{B}|\overline{u}_{x_{i}}|^{p}dx=&\sum_{i=1}^{n-1}\left(\int_{B^{+}}|u_{x_{i}}|^{p}dx+\int_{B^{-}}|(-3u(\ldots,-x_{n})+4u(\ldots,-x_{n}/2))_{x_{i}}|^{p}dx\right) \\
&+\int_{B^{+}}|u_{x_{n}}|^{p}dx+\int_{B^{-}}|3u_{x_{n}}(\ldots,-x_{n})-2u(\ldots,-x_{n}/2)|^{p}dx \\
\leq &\sum_{i=1}^{n-1}\left(\int_{B^{+}}|u_{x_{i}}|^{p}dx+3\int_{B^{-}}|u_{x_{i}}(\ldots,-x_{n})|^{p}dx+4\int_{B^{-}}|u_{x_{i}}(\ldots,-x_{n}/2)|^{p}dx\right) \\
&+\int_{B^{+}}|u_{x_{n}}|^{p}dx+3\int_{B^{-}}|u_{x_{n}}(\ldots,-x_{n})|^{p}dx+2\int_{B^{-}}|u(\ldots,-x_{n}/2)|^{p}dx \\
=&\sum_{i=1}^{n-1}\left(\int_{B^{+}}|u_{x_{i}}|^{p}dx+3\int_{B^{+}}|u_{x_{i}}|^{p}dx+2\int_{B^{+}}|u_{x_{i}}|^{p}dx\right) \\
&+\int_{B^{+}}|u_{x_{n}}|^{p}dx+3\int_{B^{+}}|u_{x_{n}}|^{p}dx+\int_{B^{+}}|u|^{p}dx \\
=&6 \sum_{i=1}^{n} \int_{B^{+}}|u_{x_{i}}|^{p}dx,
\end{align*}
where I  used the same substitutions as before.
Finally
\begin{align*}
\|\overline{u}\|_{W^{1,p}(B)} &=\left(\int_{B}|\overline{u}|^{p}dx+\sum_{i=1}^{n}\int_{B}|\overline{u}_{x_{i}}|^{p}dx\right)^{\frac{1}{p}} \\
& \leq 6^{1/p}\left(\int_{B^{+}}|u|^{p}dx+\sum_{i=1}^{n}\int_{B^{+}}|u_{x_{i}}|^{p}dx\right)^{\frac{1}{p}} \\
&=6^{1/p} \|u\|_{W^{1,p}(B^{+})}.
\end{align*}
