Assume $g_{ij}$ is Riemannian metric. Under the mean curvature flow, we have $$ \partial_t g_{ij}= -2H h_{ij} $$ where $H$ is mean curvature , and $h_{ij}$ is second fundamental form. (Forgive me talking it poorly, since this is not key point.)
Then I calculate the $\partial_t \sqrt g$ as following $$ \partial_t \sqrt g= \frac{1}{2}\frac{1}{\sqrt g} \sum_{i,j} A_{ij}\partial_t g_{ij} $$ where $g=\det(g_{ij})$ and $A_{ij}$ is algebraic cofactor of $\det(g_{ij})$. Then $$ \partial_t \sqrt g= \frac{1}{2}\frac{1}{\sqrt g} \sum_{i,j} A_{ij}\partial_t g_{ij}= -\frac{H}{\sqrt g} \sum_{i,j} A_{ij}h_{ij}=-\frac{H}{\sqrt g} \sum_{i,j} A_{ij}h_{ij}g_{ij}g^{ij}=-H^2\sqrt g $$ But I can not understand $$ -\frac{H}{\sqrt g} \sum_{i,j} A_{ij}h_{ij}=-\frac{H}{\sqrt g} \sum_{i,j} A_{ij}h_{ij}g_{ij}g^{ij}=-H^2\sqrt g \tag{1} $$ In my view, I have $$ \sum_j g_{ij}g^{ij}=1 $$ so, I think the first equation of (1) is not suitable. Besides, in the second equation, there are four $i$ and $j$, I feel disorder in the sum over $i$ and $j$.
PS: (1) is written by me, I am not sure it is right, but the result is right with the pubulic paper (they omit the process).
In writing Einstein notation, you know you screw up while you see the same lower indice. In your expression,
$$\sum_{i,j} A_{ij}h_{ij}g_{ij}g^{ij}, $$
you might probably have this in mind:
$$\sum_{i,j, k,l} A_{ij}h_{ij}g_{kl}g^{kl}.$$
So in particular you do not have
$$A_{ij}h_{ij}g_{ij}g^{ij} = (g^{ij} h_{ij}) (A_{ij} g_{ij}). $$
Instead, just use $g^{ij} = \det (g_{ij}) A_{ij}$ as suggested in the comment. Then
$$\sum_{i,j} A_{ij}h_{ij} = g \sum_{i,j} g^{ij}h_{ij} = gH. $$