Let $\Sigma$ be a compact Riemann surface. Is it possible to show that the map
$$f:\text{Div}(\Sigma)^d_+\to \text{Div}(\Sigma)^{qd}_+$$
Given by $\sum_{i} n_ix_i\mapsto \sum_{i} qn_ix_i$, has degree equal to the cardinality of the first cohomology group of the Riemann surface, with coefficients in $\mathbb F_q$? Here,
$$\text{Div}(X)_+^d$$ denotes the space of effective divisors on $X$ of degree $d$.
The way you formulated it, the answer would be no. The map $f$ is injective so the degree of the map$$\text{Div}(\Sigma)_+^d \to \text{Image}(f)$$is $1$, and if you do not take the image, then I am not sure what the degree means.
On the other hand, your hoped for conclusion is correct if you replace $\text{Div}(\Sigma)_+^d$ by its quotient of linear equivalence and suppose $d \ge g$ (the genus). By theorems of Abel and Jacobi, the quotient can be identified with the Jacobian, which is a $g$-dimensional complex torus.
