= Solution
In the regular ingoing chart, the <Killing vector field> is $k=\partial_v$, with $k^2=-V$. The normal covector to $r=b=r_+$ is $dr$; raising its index gives $\nabla r=\partial_v+V\partial_r=k$ at $r=b$. Thus $k$ is null and normal to that hypersurface, which is a <Killing horizon>. It is regular because $R^2(b)=b(b-a)>0$ when $a=r_-<b$.
The defining identity for <surface gravity> on a Killing horizon is $\nabla_a(k^2)=-2\kappa k_a$. Here $k_a=-V(dv)_a+(dr)_a$, so at the horizon this becomes $-V'(b)(dr)_a=-2\kappa(dr)_a$. Therefore
$$
\boxed{\kappa=\frac12V'(r_+)=\frac1{2r_+}.}
$$
This uses the asymptotic normalization $k^2\to-1$. A constant rescaling of $k$ rescales the surface gravity. As a check, the horizon area is $A_H=4\pi r_+(r_+-r_-)$ and $\kappa A_H/(4\pi)=(r_+-r_-)/2=M_{\rm Komar}(r_+)$. The formula is being applied only to the regular strict-inequality case.
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