Proof.
(1) For $k=0$ the first assertion holds because $\chi $ is bounded on compact sets. Let $k\geq 1$ and $1\leq j\leq d$. For a test function $\phi $ on $U$, the function $\chi \phi $ is a test function with $\partial _ j(\chi \phi )=\chi \partial _ j\phi +\phi \partial _ j\chi $, so
\[ \int \chi v\, \partial _ j\phi \, \mathrm d x=\int v\, \partial _ j(\chi \phi )\, \mathrm d x-\int v\, \partial _ j\chi \, \phi \, \mathrm d x=-\int \bigl (\chi \partial _ jv+v\partial _ j\chi \bigr )\phi \, \mathrm d x. \]
Thus $\partial _ j(\chi v)=\chi \partial _ jv+(\partial _ j\chi )v$. The functions $\partial _ jv$ and $v$ belong to $W^{k-1,p}_{\mathrm{loc}}(U)$, so by induction on $k$ both terms belong to $W^{k-1,p}_{\mathrm{loc}}(U)$; hence $\chi v\in W^{k,p}_{\mathrm{loc}}(U)$, and iterating the formula for first derivatives gives the Leibniz formula, exactly as for smooth functions.
Now let $\operatorname{Supp}\chi $ be a compact subset of $U$, and let $W$ be as in the statement. By the Leibniz formula, $\chi v$ and its derivatives of order at most $k$ vanish almost everywhere outside $\operatorname{Supp}\chi $, and there they are bounded by $C\sum _{|\beta |\leq k}|\partial ^\beta v|$, where $C$ depends on $d$, $k$ and the derivatives of $\chi $ of order at most $k$. Hence their $L^ p$ norms are at most $C\| v\| _{W^{k,p}(W)}$, which is finite when $W\Subset U$. To identify the derivatives of the extension by zero, let $\phi $ be a test function on $\mathbb R^ d$, and choose a smooth cutoff function $\zeta $ for $\operatorname{Supp}\chi $ in $U$ which equals $1$ on a neighborhood $N$ of $\operatorname{Supp}\chi $ (lemma 2.5.1(4)). Then $\zeta \phi $ is a test function on $U$, and $\partial ^\alpha (\zeta \phi )=\partial ^\alpha \phi $ on $N$. Since $\chi v$ and $\partial ^\alpha (\chi v)$ vanish almost everywhere outside $\operatorname{Supp}\chi $,
\[ \begin{split} \int \chi v\, \partial ^\alpha \phi \, \mathrm d x& =\int \chi v\, \partial ^\alpha (\zeta \phi )\, \mathrm d x\\ & =(-1)^{|\alpha |}\int \partial ^\alpha (\chi v)\, \zeta \phi \, \mathrm d x=(-1)^{|\alpha |}\int \partial ^\alpha (\chi v)\, \phi \, \mathrm d x. \end{split} \]
This proves (1).
(2) The composite $v\circ \Phi $ is defined almost everywhere: if $Z\subset U$ is a Borel null set, the change of variables formula for the diffeomorphism $\Phi ^{-1}$ gives $|\Phi ^{-1}(Z)|=\int _ Z|\det D\Phi ^{-1}(x)|\, \mathrm d x=0$. For a smooth function $w$ on $U$ and $1\leq |\alpha |\leq k$, the classical chain rule gives
\[ \partial ^\alpha (w\circ \Phi )=\sum _{1\leq |\beta |\leq |\alpha |}c_{\alpha \beta }\, (\partial ^\beta w)\circ \Phi \qquad \text{on }V, \]
with smooth functions $c_{\alpha \beta }$ on $V$ that depend only on $\Phi $: for $|\alpha |=1$ this is $\partial _ j(w\circ \Phi )=\sum _ i\partial _ j\Phi _ i\, (\partial _ iw)\circ \Phi $, and differentiating the formula once more, with the product rule and the chain rule applied to each $(\partial ^\beta w)\circ \Phi $, gives the induction step.
Assume first that $v$ vanishes almost everywhere outside the compact set $K$. By (1), applied with a smooth cutoff function equal to $1$ near $K$, the extension of $v$ by zero belongs to $W^{k,p}(\mathbb R^ d)$. Its derivatives of order at most $k$ vanish almost everywhere outside $K$, because $v$ vanishes almost everywhere on the open set $\mathbb R^ d\setminus K$ and hence so do its distributional derivatives. Choose $\delta >0$ such that the compact set $K'=\{ x:\operatorname {dist}(x,K)\leq \delta \} $ lies in $U$. For $\varepsilon <\delta $, the mollification $v_\varepsilon =v*\rho _\varepsilon $ is smooth and vanishes outside $K'$, and $\partial ^\beta v_\varepsilon =(\partial ^\beta v)*\rho _\varepsilon \to \partial ^\beta v$ in $L^ p(\mathbb R^ d)$ for $|\beta |\leq k$, by the remark preceding the lemma, applied on $\mathbb R^ d$ to a bounded open set containing $K'$, outside of which both sides vanish.
Let $J$ be the maximum of $|\det D\Phi ^{-1}|$ on $K'$, and let $M\geq 1$ bound the functions $|c_{\alpha \beta }|$ on the compact set $\Phi ^{-1}(K')$. By the change of variables formula,
\[ \int _ V|g\circ \Phi |^ p\, \mathrm d y=\int _ U|g(x)|^ p|\det D\Phi ^{-1}(x)|\, \mathrm d x\leq J\| g\| _ p^ p \]
for every measurable function $g$ on $U$ that vanishes outside $K'$. The smooth functions $w_\varepsilon =v_\varepsilon \circ \Phi $ vanish outside $\Phi ^{-1}(K')$. Applying this inequality to $v_\varepsilon -v$ and its derivatives, we find that $w_\varepsilon \to v\circ \Phi $ and, for $1\leq |\alpha |\leq k$,
\[ \partial ^\alpha w_\varepsilon \to G_\alpha =\sum _{1\leq |\beta |\leq |\alpha |}c_{\alpha \beta }\, (\partial ^\beta v)\circ \Phi \qquad \text{in }L^ p(V), \]
with $\| G_\alpha \| _ p\leq MJ^{1/p}\sum _{1\leq |\beta |\leq |\alpha |}\| \partial ^\beta v\| _ p$. For a test function $\phi $ on $V$, we obtain
\[ \begin{split} \int (v\circ \Phi )\, \partial ^\alpha \phi \, \mathrm d y& =\lim _{\varepsilon \to 0}\int w_\varepsilon \, \partial ^\alpha \phi \, \mathrm d y\\ & =(-1)^{|\alpha |}\lim _{\varepsilon \to 0}\int \partial ^\alpha w_\varepsilon \, \phi \, \mathrm d y=(-1)^{|\alpha |}\int G_\alpha \phi \, \mathrm d y. \end{split} \]
Hence $\partial ^\alpha (v\circ \Phi )=G_\alpha \in L^ p(V)$. These functions, and $v\circ \Phi $, vanish almost everywhere outside $\Phi ^{-1}(K)$, and summing the bounds over $|\alpha |\leq k$ gives $\| v\circ \Phi \| _{W^{k,p}(V)}\leq C\| v\| _{W^{k,p}(U)}$, with $C$ depending only on $k$, $p$, $J$ and $M$, that is, on $k$, $p$, $\Phi $ and $K$ once $\delta $ has been fixed in terms of $K$ and $U$.
In general, let $V'\Subset V$ be open. Choose a smooth cutoff function $\chi $ for the compact set $\Phi (\overline{V'})$ in $U$ (lemma 2.5.1(4)). By (1), $\chi v\in W^{k,p}(U)$, and it vanishes outside $\operatorname{Supp}\chi $; by the case just treated, $(\chi v)\circ \Phi \in W^{k,p}(V)$. Since $(\chi v)\circ \Phi =v\circ \Phi $ on $V'$, we obtain $v\circ \Phi \in W^{k,p}(V')$. Hence $v\circ \Phi \in W^{k,p}_{\mathrm{loc}}(V)$.
$\square$
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