= Solution
The <normalized Killing one-form> $\eta=F^{-1}V_\mu dx^\mu$ is a <closed differential form>. The <Poincaré lemma> gives a local scalar potential $\phi$ with $\eta=d\phi$, hence $V_\mu=F\partial_\mu\phi$. Choose the adapted chart of part (i), so $V=\partial_0$. Then
$$
\partial_0\phi=\eta(V)=1,\qquad
\phi=x^0+f(x^1,x^2,x^3),\qquad
\partial_i f=\frac{g_{0i}}{g_{00}}.
$$
The <closed differential form> condition supplies the integrability of these spatial derivatives. For a fixed $g$ and $V$, $f$ is determined locally up to an additive constant; its derivatives cannot be chosen freely despite the wording in the paper.
Use $\bar x^0=\phi$, $\bar x^i=x^i$. The <Jacobian determinant> is one, so this is a local coordinate change. Holding $\bar x^i$ fixed gives $V=\partial_{\bar0}$, while its metric-dual <one-form> is $v=F\,d\bar x^0$. Thus $\bar g_{00}=F<0$ and $\bar g_{0i}=0$. The <Killing equation> still holds, so $\partial_{\bar0}\bar g_{\mu\nu}=0$. \b[The metric is locally static]:
$$
\boxed{\bar g_{0i}=0,\qquad\partial_{\bar0}\bar g_{\mu\nu}=0.}
$$
A <closed differential form> need not be globally exact; neither a global potential nor a global static chart follows without additional topological assumptions.
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