Let's first look at the proof of ordinary Grinberg's theorem.
Suppose that $G$ has $n$ vertices. Let $G_1$ be the subgraph of $G$ formed by deleting all edges outside $C$. Then $G_1$ also has $n$ vertices. The sum $\sum k f_k$ counts every edge of $G_1$ twice, except that it counts the $n$ edges of $C$ only once, so $G_1$ has $\frac12(n + \sum k f_k)$ edges. The sum $\sum f_k$ counts every face of $G_1$ except the external face, so $G$ has $1 + \sum f_k$ faces. Therefore Euler's formula says
$$
n - \frac12\left(n + \sum_{k\ge 3} k f_k\right) + 1 + \sum_{k \ge 3} f_k = \color{red}{2}
$$
and by rearranging, we get
$$
\sum_{k \ge 3} \left(k-2\right)f_k = n + 2 - 2\cdot\color{red}{2}.
$$
Let $G_2$ be the subgraph of $G$ formed by deleting all edges of $G$ inside $C$, and we get the same expression for the sum with $g_k$. Taking the difference gives us Grinberg's theorem.
In the case of a higher-genus embedding, the first complication we encounter is that Euler's formula will replace $\color{red}{2}$ in the argument above by $\color{red}{2 - 2g(G_1)}$, where $g(G_1)$ is the genus of $G_1$. Specifically, this is the genus of the particular embedding of $G_1$ we're looking at above, assuming that all faces are actually, topologically, disks. This may not be the same as $g(G)$.
Another way to think of $g(G_1)$ is that we take our embedding of $G$ in a surface, cut along $C$ and take the piece we called the "inside", and glue a disk along the boundary. Then $g(G_1)$ is the genus of the resulting surface. Similarly, $g(G_2)$ is the genus of the surface we get in this way, but starting from the piece we called the "outside".
Anyway, once we've figured out what $g(G_1)$ is, then we have
$$
\sum_{k\ge 3}(k-2)f_k = n - 2 + 4 g(G_1)
$$
and therefore
$$
\sum_{k\ge 3}(k-2)(f_k - g_k) = 4 g(G_1) - 4g(G_2).
$$
Grinberg's theorem still holds, but with an error term when one piece of the surface has a different genus from the other.
There's another complication, which is that it's possible for none of this to make any sense. In higher-genus surfaces, it's possible that cutting along $C$ does not divide the surface into two pieces. In this case, there's no interesting statement to be made, because there's no way to divide faces of $G$ into "inside" and "outside" faces.
In the specific case of the double torus, there's three cases, illustrated in the gluing diagram below:
If $C$ follows the orange loop, then cutting along $C$ and gluing disks gives us a sphere and a double torus, so the embeddings of $G_1$ and $G_2$ have genus $0$ and $2$ (in some order). Grinberg's theorem will hold with an error term of $8$.
If $C$ follows the purple loop, then cutting along $C$ and gluing disks divides the double torus into two single tori, so the embeddings of $G_1$ and $G_2$ both have genus $1$. Grinberg's theorem will hold with no error term.
If $C$ follows the blue loop, then cutting along $C$ doesn't leave two pieces: instead, we get a single manifold whose boundary is two circles. So we can't divide the faces of $G$ into two types, and Grinberg's theorem has nothing meaningful to say.
