 Solution

Locally write the null hypersurface as $S=0$ with $dS\ne 0$, and extend its normal as $n_a={\nabla }_{a}S$. On the hypersurface, $n^a{n}_a=0$, so $n^a{\nabla }_{a}S=n^2=0$: the raised normal is itself tangent. The Levi-Civita connection and symmetry of the scalar Hessian giveAlthough $n^2$ vanishes on the hypersurface, it need not vanish off it. Its tangential derivatives vanish, so ${\nabla }_a(n^2)$ must be proportional to the normal $n_a$. Hence ${\nabla }_{n}n=\kappa n$ on the hypersurface for some smooth local coefficient $\kappa$. The normal's integral curves of a vector field remain in the hypersurface and satisfy the nonaffine geodesic equation. Rescale to $k=hn$, where $n(\log h)=-\kappa$ along each generator. Direct substitution gives ${\nabla }_{k}k=0$. Thus the normal generates null geodesics within the hypersurface, with an affine parameter after this rescaling. This proves the null hypersurface normal generates null geodesics property without incorrectly setting a transverse derivative of $n^2$ to zero.