= Solution
Use $\hbar=c=k_B=1$ and normalize the time-translation <Killing vector field> at infinity. The simple outer zero and positivity on its exterior imply $b=V'(r_0)>0$. Near the horizon, write $x=r-r_0$; then $V=bx+O(x^2)$. The <Wick rotation> $t=-it_E$ gives the radial Euclidean metric
$$
ds_E^2=V\,dt_E^2+\frac{dr^2}{V}+r^2d\Omega^2.
$$
Define $\rho=2\sqrt{x/b}$, so $x=b\rho^2/4$. Its leading near-horizon form is
$$
ds_E^2=d\rho^2+\rho^2\left(\frac b2dt_E\right)^2+r_0^2d\Omega^2+O(\rho^2)d\rho^2+O(\rho^4)dt_E^2+O(\rho^2)d\Omega^2.
$$
Thus $bt_E/2$ is a polar angle. Smoothness at the origin requires period $2\pi$, rather than a cone or a multiple cover with a branched origin. The <Euclidean black-hole regularity condition> is
$$
\beta=\frac{4\pi}{b}.
$$
A thermal quantum state has imaginary-time period $\beta=1/T$. Since $V\to1$, this time coordinate is normalized to the clocks at infinity, and the <Hawking temperature> is
$$
\boxed{T=\frac{V'(r_0)}{4\pi}.}
$$
A static clock at radius $r>r_0$ measures the redshifted local temperature $T/\sqrt{V(r)}$; that local quantity is not the temperature requested with the asymptotic normalization.
The static coordinates are singular on the horizon, so calculate its <surface gravity> in a regular ingoing chart. With $v=t+\int dr/V$, the metric and the same normalized <Killing vector field> are
$$
ds^2=-V\,dv^2+2\,dv\,dr+r^2d\Omega^2,\qquad k=\partial_v.
$$
The inverse radial block has $g^{vr}=1$, $g^{rr}=V$, $g^{vv}=0$. Direct calculation gives
$$
\Gamma^v{}_{vv}=\frac{V'}2,\qquad\Gamma^r{}_{vv}=\frac{VV'}2.
$$
Since $k$ has constant coordinate components, its acceleration is
$$
k^a\nabla_a k^b=\frac{V'}2(\partial_v)^b+\frac{VV'}2(\partial_r)^b.
$$
On the future horizon, $V=0$ and $k$ is null, so the defining equation gives $\kappa=b/2$. In the outgoing chart regular on the past branch, the radial cross-term is $-2\,du\,dr$ and the same calculation gives $\kappa=-b/2$ for $k=\partial_u$. Hence the branch-independent physical statement is
$$
\boxed{|\kappa|=\frac{V'(r_0)}2=2\pi T.}
$$
This proves the <Euclidean temperature and signed horizon surface gravity> relation directly. Rescaling the <Killing vector field> would rescale its <surface gravity>; the condition at infinity fixes that otherwise arbitrary normalization.
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