Hardness of a problem which is the sum of two NP-Hard problems
Nothing.
Lower bound: Suppose $h(x) = -f(x)$. Then $\sum_x g(x) = 0$, which is trivial to compute. If $\sum_x f(x)$ is NP-hard to compute, then $\sum_x h(x)$ will be too, but $\sum_x g(x)$ will be easy to compute.
Upper bound: Suppose $h(x) = f(x)$. Then $\sum_x g(x) = 2 \sum_x f(x)$, which is as hard as computing $\sum_x f(x)$, which is assumed to be NP-hard. Since NP-hard means (roughly speaking) NP-complete or harder, there is no upper bound on the complexity; it could be arbitrarily hard to compute $\sum_x g(x)$.