Surface gravity of Kerr black hole

The calculation can be done in this coordinate system just fine, even though it doesn't extend across the horizon. Surface gravities are very commonly computed in coordinate systems which go bad at the horizon. For example, the surface gravity of Schwarzschild

$ds^2 = -f dt^2 + f^{-1} dr^2 + r^2 d\Omega_2^2, \qquad f= 1- \frac{r_+}{r},$

is easily found to be $\kappa = \frac{f'}{2} = \frac{1}{2r_+}$.

I think your problem is that you are evaluating quantities at the horizon before taking derivatives. It's important to first take derivatives, and then to evaluate at the horizon.

You are right that the $(\Omega_H-\omega)^2$ term doesn't contribute. This is because it is the square of something that vanishes at the horizon: when you take the derivative, there remains a vanishing factor. As for the other term, since $\Delta$ vanishes at the horizon, this term vanishes except when the derivative hits $\Delta$. This yields the last formula you wrote.

$$\nabla_\nu(-\chi^\mu\chi_\mu) =\partial_\nu(-\chi^\mu\chi_\mu)\\ =\frac{\rho^2 }{\Sigma}\partial_\nu \Delta + \Delta \partial_\nu(\frac{\rho^2 }{\Sigma})-(\Omega_H -\omega)^2 \partial_\nu(\frac{\Sigma\sin^2{\theta}}{\rho^2})$$

Now use the horizon condition you will get $$\nabla_\nu(-\chi^\mu\chi_\mu)=\frac{\rho^2 }{\Sigma}\partial_\nu \Delta $$

Since $\chi^\mu$ is null on horizon and a null vector is normal to itself so $\chi^\mu$ must be proportional to the normal of the horizon. A constant r surface has a normal $\partial_\mu{r}$. So $$\chi_{\mu}=C\partial_\mu{r}$$ Now our job is to find C. It can be found easily

$$g^{\mu\nu}\chi_{\mu}\chi_{nu}=C^2g^{\mu\nu}\partial_\mu{r}\partial_\nu{r}\\ =C^2g^{rr}$$

So $$C^2=\frac{\chi^{\mu}\chi_{\mu}}{g^{rr}}$$

So after the algebra is done take the horizon limit and you will find C. The rest are just few lines algebra.