 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 $-V^{\prime }(b)(dr{)}_a=-2\kappa (dr{)}_a$. ThereforeThis 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 $\kappa A_H/(4\pi )=(r_{+}-r_{-})/2=M_{\mathrm{Komar}}(r_{+})$. The formula is being applied only to the regular strict-inequality case.