Higher Genus Grinberg's Theorem Grinberg's Theorem is formulated like the following:

Let $G$ be a finite planar graph with a Hamiltonian cycle $C$, with a fixed planar embedding. Denote by $ƒ_k$ and $g_k$ the number of $k$-gonal faces of the embedding that are inside and outside of $C$, respectively. Then $$\sum _{{k\geq 3}}(k-2)(f_{k}-g_{k})=0.$$

The proof is an easy consequence of Euler's formula.
How does Grinberg's Theorem look like if we look at surfaces with higher genus
$$    {\displaystyle 2-2g =V-E+F}$$
Since in the planar case, the boundary edges of the faces inside and outside $C$ have to match, I don't expect Grinberg's Theorem to change.
But is that right?
Update I found Grinberg’s Criterion by Gunnar BRINKMANN and Carol T. ZAMFIRESCU, which "generalize(s) Grinberg’s hamiltonicity criterion for planar graphs.", but I get lost by "the many parameters just defined" before the main theorem.
Can anybody help me to understand it?
If it helps, I'm specially interested in the case of strongly embedded bicubic graphs on a double torus made up of hexagons and octagons only...
 A: 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.
