[proofplan]
For each parameter value $s$, the scalar mean curvature $H_s$ of the hypersurface $F_s(\Sigma)$ vanishes because the hypersurface is minimal. We differentiate this identity at $s=0$. The [first variation formula](/theorems/2728) for scalar mean curvature says that the derivative equals the tangential derivative of $H_0$ along the tangential part of the variation minus $L\phi$; since $H_0=0$, the tangential term vanishes. Therefore $0=-L\phi$, which gives the Jacobi field equation.
[/proofplan]
[step:Define the scalar mean curvature functions along the variation]
Let $g$ denote the Riemannian metric on the ambient manifold $M$. Since the family is defined for $s$ near $0$, choose a number $\varepsilon>0$ such that $F_s:\Sigma\to M$ is a smooth immersion with two-sided minimal image for every $s\in(-\varepsilon,\varepsilon)$.
For each $s \in (-\varepsilon,\varepsilon)$, let $g_s := F_s^*g$ be the Riemannian metric on $\Sigma$ induced by $F_s$. Let $A_s$ denote the second fundamental form of the immersion $F_s$ with respect to the chosen unit normal field $\nu_s$, and let $H_s: \Sigma \to \mathbb{R}$ denote the scalar mean curvature, defined as the trace of $A_s$ with respect to $g_s$.
Since each image $F_s(\Sigma)$ is minimal, its scalar mean curvature is identically zero. Hence
\begin{align*}
H_s(p)=0
\end{align*}
for every $s \in (-\varepsilon,\varepsilon)$ and every $p \in \Sigma$.
[/step]
[step:Decompose the variational vector field into tangential and normal parts]
Let $V: \Sigma \to TM$ be the variational vector field along $F_0$. Since $F_0$ is an immersion and $\nu_0$ is a unit normal field along $F_0$, every vector $V(p) \in T_{F_0(p)}M$ decomposes uniquely into a tangential part and a normal part:
\begin{align*}
V(p)=d(F_0)_p(X(p))+\phi(p)\nu_0(p),
\end{align*}
where $X: \Sigma \to T\Sigma$ is a smooth vector field on $\Sigma$ and $\phi: \Sigma \to \mathbb{R}$ is the smooth function from the statement. The smoothness of $X$ and $\phi$ follows from the smooth orthogonal splitting
\begin{align*}
F_0^*TM = dF_0(T\Sigma) \oplus \operatorname{span}(\nu_0).
\end{align*}
[/step]
[step:Differentiate the zero mean curvature identity]
Fix $p \in \Sigma$. Define the smooth function $h_p: (-\varepsilon,\varepsilon) \to \mathbb{R}$ by $h_p(s)=H_s(p)$. This function is identically zero. Therefore
\begin{align*}
\left.\frac{d}{ds}\right|_{s=0}H_s(p)=0.
\end{align*}
The [First Variation Formula for Scalar Mean Curvature](/theorems/first-variation-formula-for-scalar-mean-curvature) under the variation field $V=dF_0(X)+\phi\nu_0$ gives
\begin{align*}
\left.\frac{d}{ds}\right|_{s=0}H_s = X(H_0)-L\phi.
\end{align*}
Here $X(H_0): \Sigma \to \mathbb{R}$ is the derivative of the function $H_0$ in the direction of the vector field $X$, and $L$ is the Jacobi operator
\begin{align*}
L\phi=\Delta_\Sigma\phi+\left(|A|^2+\operatorname{Ric}_M(\nu_0,\nu_0)\right)\phi.
\end{align*}
The hypotheses of this linearisation formula are satisfied: the variation $F:(-\varepsilon,\varepsilon)\times\Sigma\to M$ is smooth, each $F_s$ is a smooth immersion, the unit normals $\nu_s$ are chosen smoothly near $s=0$, and the variation field has the decomposition $V=dF_0(X)+\phi\nu_0$ with $X\in\mathfrak{X}(\Sigma)$ and $\phi\in C^\infty(\Sigma)$. This is the standard linearisation formula for scalar mean curvature with the sign convention used in the statement.
Since $H_0=0$ as a function on $\Sigma$, its derivative in every tangential direction vanishes:
\begin{align*}
X(H_0)=0.
\end{align*}
Combining the two identities gives
\begin{align*}
0
=
\left.\frac{d}{ds}\right|_{s=0}H_s
=
-L\phi.
\end{align*}
Hence $L\phi=0$ on $\Sigma$.
[guided]
The point of this step is to isolate the only part of the variation that changes the hypersurface geometrically to first order. Define the variational vector field $V:\Sigma\to TM$ along $F_0$ by
\begin{align*}
V(p)=\left.\frac{\partial}{\partial s}\right|_{s=0}F_s(p).
\end{align*}
For each $p\in\Sigma$, the vector $V(p)$ lies in $T_{F_0(p)}M$. Since $F_0$ is an immersion and $\nu_0$ is a unit normal field along $F_0$, the orthogonal splitting
\begin{align*}
F_0^*TM=dF_0(T\Sigma)\oplus\operatorname{span}(\nu_0)
\end{align*}
gives a unique decomposition
\begin{align*}
V(p)=d(F_0)_p(X(p))+\phi(p)\nu_0(p),
\end{align*}
where $X\in\mathfrak{X}(\Sigma)$ is a smooth vector field and $\phi\in C^\infty(\Sigma)$ is the smooth normal speed from the statement. The tangential component changes the parametrisation of the hypersurface; the normal component moves the hypersurface through nearby hypersurfaces.
For each fixed $p \in \Sigma$, define the smooth function $h_p: (-\varepsilon,\varepsilon) \to \mathbb{R}$ by $h_p(s)=H_s(p)$. Since every $F_s(\Sigma)$ is minimal, $H_s(p)=0$ for every $s$, so differentiating the constant function $h_p$ gives
\begin{align*}
\left.\frac{d}{ds}\right|_{s=0}H_s(p)=0.
\end{align*}
Now we use the [First Variation Formula for Scalar Mean Curvature](/theorems/first-variation-formula-for-scalar-mean-curvature). Its hypotheses are exactly the regularity and two-sidedness data available here: the map $F:(-\varepsilon,\varepsilon)\times\Sigma\to M$ is smooth, each slice $F_s:\Sigma\to M$ is a smooth immersion, the normal field $\nu_s$ is chosen smoothly near $s=0$, and the variational vector field decomposes as $V=dF_0(X)+\phi\nu_0$ with $X\in\mathfrak{X}(\Sigma)$ and $\phi\in C^\infty(\Sigma)$. Therefore the derivative of scalar mean curvature at $s=0$ is
\begin{align*}
\left.\frac{d}{ds}\right|_{s=0}H_s = X(H_0)-L\phi.
\end{align*}
The term $X(H_0)$ is the contribution of the tangential reparametrisation, while the term $-L\phi$ is the contribution of the normal deformation. With the sign convention in the statement, the Jacobi operator is
\begin{align*}
L\phi
=
\Delta_\Sigma\phi+
\left(|A|^2+\operatorname{Ric}_M(\nu_0,\nu_0)\right)\phi.
\end{align*}
The tangential term disappears because the base hypersurface is minimal. More precisely, $H_0$ is the zero function on $\Sigma$, and therefore the derivative of $H_0$ in the direction of any vector field $X$ is also the zero function:
\begin{align*}
X(H_0)=0.
\end{align*}
Substituting this into the variation formula yields
\begin{align*}
0
=
\left.\frac{d}{ds}\right|_{s=0}H_s
=
X(H_0)-L\phi
=
-L\phi.
\end{align*}
Thus
\begin{align*}
L\phi=0.
\end{align*}
This proves that the normal speed $\phi$ of any smooth variation through minimal hypersurfaces lies in the kernel of the Jacobi operator.
[/guided]
[/step]
[step:Conclude the Jacobi field equation]
The previous step shows that for every point $p \in \Sigma$,
\begin{align*}
(L\phi)(p)=0.
\end{align*}
Therefore $L\phi=0$ as a smooth function on $\Sigma$, which is the claimed Jacobi field equation for the normal part of the variation.
[/step]
