A question on Levi-Civita connection and a fixed hyper surface Suppose $(M,g)$ is a three dimensional smooth compact simply connected Riemannian manifold with boundary and suppose that $\Sigma$ is a smooth simply connected hypersurface in $M$ with a smooth boundary $\partial \Sigma \subset \partial M$. Let $D_X Y$ denote the Levi-Civita connection on $(M,g)$. Let $Z$ be a smooth vector field on $\Sigma$ and let $f:\Sigma \to \mathbb R$ be a smooth function on $\Sigma$. Does there exist a smooth function $\phi$ in $M$ such that $\Sigma=\{\phi=0\}$ and such that
$$ (D_{\nabla \phi} \nabla \phi)\big|_{\Sigma} = f\, Z.$$
Note that the left hand side is just the restriction of $D_{\nabla \phi}\nabla \phi$ to $\Sigma$ and that $\nabla \phi$ is the gradient of $\phi$ with respect to $g$.
If the answer is no, would it make a difference if the latter condition is replaced with
$$(D_{\nabla \phi} \nabla \phi)\big|_{\Sigma} - f\, Z \in \textrm{Span}\{(\nabla \phi)|_{\Sigma}\}$$
 A: Not a full answer, but there definitely should be some integrability constraint.
Take the simplest case where $M = \mathbb{R}^3$ (the boundary is unimportant for the discussion here) and $\Sigma$ is the $x$-$y$ plane.
If $\phi$ is a defining function of $\Sigma$, then restricted to $\Sigma$ we have $\nabla \phi|_{\Sigma} = \eta \partial_z$ for some function $\eta: \Sigma \to \mathbb{R}$.
Symmetry of the Hessian requires then
$$ \nabla^2 \phi|_{\Sigma} = \begin{pmatrix}
0 & 0 & \partial_x \eta \\
0 & 0 & \partial_y \eta \\
\partial_x \eta & \partial_y \eta & * \end{pmatrix} $$
This shows that
$$ D_{\nabla\phi} \nabla\phi = \eta (\partial_x \eta, \partial_y \eta, *) $$
So a necessary condition for your equation to hold (setting $f \equiv 1$ since as you agreed it is unimportant) in the flat case is that $Z$ is the gradient of some scalar function on $\Sigma$, and that allowing there to be normal components (as in your modified question) has no impact.

The conclusion is unchanged in the Lorentzian setup, if you require $\Sigma$ to be non-degenerate. In the case where $\Sigma$ is null, however, since $\nabla\phi$ now lies in the tangent space of $\Sigma$ and $\Sigma$ is two dimensional, it may in fact be possible for the second variant to hold (but I don't have the energy to run through the analysis now).
