= Solution
Set $r_H=2m$. Since $f'(r_H)=2\kappa$, the integrand defining the <tortoise coordinate> has a simple pole:
$$
r_* =\frac1{2\kappa}\log|r-r_H|+h(r),
$$
where $h$ is analytic near $r_H$. Define the <Kruskal–Szekeres coordinates> in the right exterior by $U=-e^{-\kappa u}$ and $V=e^{\kappa v}$. Then
$$
-UV=e^{2\kappa r_*}=(r-r_H)e^{2\kappa h(r)}.
$$
The right-hand side has positive nonzero derivative at $r_H$. Continuing this relation analytically, rather than retaining an absolute value, makes $r$ an analytic function of $UV$ through both signs. The <Kruskal extension across a simple static horizon> has metric
$$
ds^2=\frac{f(r)}{\kappa^2UV}\,dU\,dV+R(r)^2d\Omega^2.
$$
Its radial coefficient tends to $-2e^{-2\kappa h(r_H)}/\kappa$, finite and nonzero, while $R(r_H)^2=4m^2\cosh\alpha$. Therefore both null surfaces $U=0$ and $V=0$ are regular horizon branches, intersecting at a smooth two-sphere $U=V=0$.
The stationary <Killing vector field> becomes
$$
\boxed{K=\kappa\bigl(V\partial_V-U\partial_U\bigr).}
$$
It is null, tangent and normal on each horizon branch and vanishes on their intersection. This proves that $r=2m$ is a <bifurcate Killing horizon>, with <bifurcation surface> of area $\boxed{A_H=16\pi m^2\cosh\alpha}$. The exponential definitions specify one exterior; allowing all signs of $U,V$ supplies the remaining quadrants of the extension.
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