As picture below, I want to show the volume enclosed by $M_t$ is constant. Because $$ V(t)=\frac{1}{n}\int_{M_t} F(x,t)\cdot \nu(x,t) dS $$ And $\partial_t\sqrt g=-H(H-h)\sqrt g$, so $$ \frac{dV(t)}{dt}=\frac{1}{n}\int_{M_t} [\partial_tF(x,t)\cdot \nu(x,t) +F(x,t)\cdot \partial_t\nu(x,t)-H(H-h)F(x,t)\cdot \nu(x,t)] dS \\ =\frac{1}{n}\int_{M_t} [(h(t)-H(x,t))\nu(x,t)\cdot \nu(x,t) +F(x,t)\cdot \partial_t\nu(x,t)-H(H-h)F(x,t)\cdot \nu(x,t)] dS \\ =\frac{1}{n}\int_{M_t}[F(x,t)\cdot \partial_t\nu(x,t)-H(H-h)F(x,t)\cdot \nu(x,t)] dS $$ Then, to show $\int_{M_t}[F(x,t)\cdot \partial_t\nu(x,t)-H(H-h)F(x,t)\cdot \nu(x,t)] dS =0$
Because $\nu(x,t)$ is outer unit normal vector. We have $\langle\nu(x,t),\nu(x,t)\rangle=1$. Then $$ \langle \partial_t\nu(x,t),\nu(x,t) \rangle=0 $$ Besides, $\{ \partial_iF \}$ is a basis of $T_xM$, so we can assume $$ \partial_t\nu (x,t)=v^i\partial_iF $$ Then, we have $$ \partial_t\nu= \langle \partial_t\nu, \partial_jF\rangle \partial_j F g^{ij} $$ Because $\langle \nu, \partial_jF\rangle = 0$, we have $\langle \partial_t\nu, \partial_jF\rangle =-\langle \nu, \partial_t\partial_jF\rangle$. Then $$ \partial_t\nu=\langle \nu,\partial_i(H\nu) \rangle \partial_jF g^{ij}=\partial_iH \partial_jFg^{ij}=\nabla H ~~~~~\text{gradient on M} $$ So $$ F(x,t)\cdot \partial_t\nu(x,t)-H(H-h)F(x,t)\cdot \nu(x,t) \\ =F\cdot\nabla H-H(H-h)F(x,t)\cdot \nu(x,t) $$ Then, I get stuck.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This is very simple, so I'm not going to try to read your long calculation. If you move a surface in the outwards normal direction at speed $f(x)$, then the enclosed volume changes by $dV/dt = \int f dA$. Since $h$ is defined as the average of $H$, $h-H$ has average $0$, and thus $dV/dt = \int (h - H) dA = 0$.