= Solution
Locally write the <null hypersurface> as $S=0$ with $dS\ne0$, and extend its normal as $n_a=\nabla_aS$. On the hypersurface, $n^an_a=0$, so $n^a\nabla_aS=n^2=0$: the raised normal is itself tangent. The <Levi-Civita connection> and symmetry of the scalar Hessian give
$$
n^b\nabla_bn_a=n^b\nabla_an_b=\frac12\nabla_a(n^2).
$$
Although $n^2$ 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 $\nabla_nn=\kappa n$ 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 $\nabla_kk=0$. Thus \b[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.
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