[guided]We verify the variation formula at a fixed point $p\in\Sigma$. The calculation below uses only the initial velocity field $V=\phi\nu$ at $t=0$: all differentiated quantities are evaluated at $t=0$, and the possible $t$-dependence of the speed function away from $t=0$ does not enter the linearization of $H_t$. Choose a local tangent frame $e_1,\dots,e_m$ near $p$ that is orthonormal at $p$ and satisfies $\nabla^\Sigma_{e_i}e_j(p)=0$ for all $1\le i,j\le m$. Extend the frame along the variation by pushforward: $e_i(t)=dF_t(e_i)$. The pushed-forward frame is usually not orthonormal for $t\ne0$, so we must differentiate the inverse induced metric as well as the second fundamental form. This inverse-metric variation is not negligible; it is one of the terms that produces the $-|A|^2\phi$ contribution.
Let $g_{ij}(t)=g(e_i(t),e_j(t))$ denote the induced metric matrix on $\Omega_t$ in the moving frame, and let $g^{ij}(t)$ denote the entries of its inverse matrix. Let
\begin{align*}
h_{ij}(t)=g(\nabla^M_{e_i(t)}\nu_t,e_j(t))
\end{align*}
denote the components of the second fundamental form of $\Omega_t$ with respect to $\nu_t$. The scalar mean curvature is therefore
\begin{align*}
H_t=\sum_{i,j=1}^m g^{ij}(t)h_{ij}(t).
\end{align*}
At $t=0$ and $p$, $g^{ij}(0)=\delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta.
The normal velocity is $V=\phi\nu$. Differentiating the orthogonality relation $g(\nu_t,e_i(t))=0$ at $t=0$ gives
\begin{align*}
g(\nabla^M_V\nu,e_i)+g(\nu,\nabla^M_{e_i}V)=0.
\end{align*}
Since $\nabla^M_{e_i}V=e_i(\phi)\nu+\phi\nabla^M_{e_i}\nu$ and $g(\nabla^M_{e_i}\nu,\nu)=0$, this becomes
\begin{align*}
g(\nabla^M_V\nu,e_i)=-e_i(\phi).
\end{align*}
Differentiating $g(\nu_t,\nu_t)=1$ also gives $g(\nabla^M_V\nu,\nu)=0$, so $\nabla^M_V\nu$ is tangent. Hence
\begin{align*}
\nabla^M_V\nu=-\nabla^\Sigma\phi.
\end{align*}
Now we differentiate the two factors in $H_t=\sum_{i,j=1}^m g^{ij}(t)h_{ij}(t)$. First, differentiating the induced metric gives
\begin{align*}
g'_{ij}(0)=g(\nabla^M_{e_i}V,e_j)+g(e_i,\nabla^M_{e_j}V).
\end{align*}
Using $V=\phi\nu$, $g(\nu,e_i)=0$, and $A_{ij}=g(\nabla^M_{e_i}\nu,e_j)$, this becomes
\begin{align*}
g'_{ij}(0)=2\phi A_{ij}.
\end{align*}
Since the derivative of an inverse matrix satisfies $(g^{ij})'(0)=-g'_{ij}(0)$ at a point where $g_{ij}(0)=\delta_{ij}$, we obtain
\begin{align*}
(g^{ij})'(0)=-2\phi A_{ij}.
\end{align*}
Consequently the inverse-metric part contributes
\begin{align*}
\sum_{i,j=1}^m (g^{ij})'(0)h_{ij}(0)=-2\phi |A|^2.
\end{align*}
This is the term that would be missed if the pushed-forward frame were treated as orthonormal away from $t=0$.
Next we differentiate $h_{ij}(t)$. Work on the swept-out map $F:(-\varepsilon,\varepsilon)\times\Omega\to M$ and extend $V$ to the velocity field $\partial_tF$ along this image, while extending each $e_i$ by the pushed-forward field $e_i(t)=dF_t(e_i)$. These fields are $F$-related to the coordinate vector field $\partial_t$ and to the time-independent vector field $e_i$ on $\Omega$, respectively, so their Lie bracket along the variation satisfies $[V,e_i(t)]=0$. Using the curvature convention
\begin{align*}
R^M(X,Y)Z=\nabla^M_X\nabla^M_YZ-\nabla^M_Y\nabla^M_XZ-\nabla^M_{[X,Y]}Z,
\end{align*}
we get
\begin{align*}
h'_{ij}(0)
=
g(\nabla^M_{e_i}\nabla^M_V\nu,e_j)
+
g(R^M(V,e_i)\nu,e_j)
+
g(\nabla^M_{e_i}\nu,\nabla^M_{e_j}V).
\end{align*}
Tracing with $\delta_{ij}$ gives three terms. Since $\nabla^M_V\nu=-\nabla^\Sigma\phi$ is tangent and $\nabla^\Sigma_{e_i}e_i(p)=0$, the first traced term is
\begin{align*}
\sum_{i=1}^m g(\nabla^M_{e_i}\nabla^M_V\nu,e_i)=-\Delta_\Sigma\phi.
\end{align*}
For the curvature term, $V=\phi\nu$. With the Ricci convention $\operatorname{Ric}_M(\nu,\nu)=\sum_{i=1}^m g(R^M(e_i,\nu)\nu,e_i)$ and the skew-symmetry of $R^M$ in the first two slots, we obtain
\begin{align*}
\sum_{i=1}^m g(R^M(V,e_i)\nu,e_i)=-\phi\operatorname{Ric}_M(\nu,\nu).
\end{align*}
For the last traced term, the identity $\nabla^M_{e_i}V=e_i(\phi)\nu+\phi\nabla^M_{e_i}\nu$ and the tangency of $\nabla^M_{e_i}\nu$ give
\begin{align*}
\sum_{i=1}^m g(\nabla^M_{e_i}\nu,\nabla^M_{e_i}V)=\phi |A|^2.
\end{align*}
Therefore the differentiated second-fundamental-form part contributes
\begin{align*}
\sum_{i=1}^m h'_{ii}(0)=-\Delta_\Sigma\phi-\phi\operatorname{Ric}_M(\nu,\nu)+\phi |A|^2.
\end{align*}
Combining this with the inverse-metric contribution $-2\phi |A|^2$ yields
\begin{align*}
\left.\frac{\partial H_t}{\partial t}\right|_{t=0}
=
-\Delta_\Sigma \phi-
\left(|A|^2+\operatorname{Ric}_M(\nu,\nu)\right)\phi.
\end{align*}[/guided]
