Let $(M^{n+1},\bar{g})$ be a space form (i.e. a Riemannian manifold with constant sectional curvature, says $K\in\mathbb{R}$). Let $\Sigma$ be an orientable hypersurface in $M$, with induced metric $g$ and unit normal vector field $\nu$.
Let $X$ be a Killing vector field on $M$ (i.e. $\mathcal{L}_X\bar{g}=0$). It is known that if $\Sigma$ has constant mean curvature, then the function \begin{align} f:=\langle X,\nu\rangle \end{align} (I will often use $\langle,\rangle$ to denote $\bar{g}$) satisfies the Jacobi equation: \begin{align} \Delta_{\Sigma}f+\big(|A|^2+\overline{Ric}(\nu,\nu)\big)f=0 \end{align} where $\Delta_{\Sigma}$ is the Laplacian on $\Sigma$ in the induced metric, $A=(h_{ij})$ is the second fundamental form of $\Sigma$, and $\overline{Ric}$ is the Ricci curvature of $(M,\bar{g})$.
This result seems to be quite standard. However I fail to find any reference that gives a proof on it, so I try to prove it by myself as an exercise. Fix a point $p\in\Sigma$ and choose a local orthonormal frame $\{e_i\}_{i=1}^n$ around $p$ such that
\begin{align}
\nabla_{e_i}e_j\big|_p=0 & & (1)
\end{align}
where $\nabla$ is the Levi-Civita connection of $(\Sigma,g)$ and I will denote $D$ the Levi-Civita connection of $(M,\bar{g})$. Then we will have, at point $p$,
\begin{align}
\Delta_{\Sigma}f=\sum_{i=1}^n\langle D_{e_i}(D_{e_i}X),\nu\rangle
+2\sum_{i=1}^n\langle D_{e_i}X,D_{e_i}\nu\rangle
+\sum_{i=1}^n\langle X,D_{e_i}(D_{e_i}\nu)\rangle
\end{align}
I can show that the 2nd term on R.H.S. vanishes and the 3rd term gives $\langle X,\nabla H\rangle-|A|^2\langle X,\nu\rangle$, so that with $H\equiv const$ and rearrange the equation, we get that at $p$,
\begin{align}
\Delta_{\Sigma}f+|A|^2f=\sum_{i=1}^n\langle D_{e_i}(D_{e_i}X),\nu\rangle & & (2)
\end{align}
Now I would like to know how to obtain $-\overline{Ric}(\nu,\nu)\langle X,\nu\rangle$ from the 1st term (the R.H.S. of (2)). Of course, since $(M,\bar{g})$ is a space form, this term is also equal to $-\overline{Ric}(X,\nu)$.
It would perhaps be too long to include my full computation here (surely, I am ready to provide them per request). In short, first using the Killing property of $X$ and (1), I am able to compute that at $p$, \begin{align} \sum_{i=1}^n\langle D_{e_i}(D_{e_i}X),\nu\rangle=-\sum_{i=1}^n\langle D^2_{e_i,\nu}X,e_i\rangle \end{align} then we can use the definition of Riemann curvature tensor to yield an $\overline{Rm}$-term, which then yields $\overline{Ric}(X,\nu)$ after taking trace. However, the other terms seems to give an extra term, so that in summary, I get \begin{align} -\overline{Ric}(X,\nu) +\sum_{i=1}^n\langle D_{[\nu,e_i]}X,e_i\rangle \end{align} where $[,]$ is the Lie bracket on $(M,\bar{g})$. I wonder how do we get rid of this extra last term.
Any hint, comment and answer are welcomed and greatly appreciated. Any reference which contain a proof of it will also be great.