If I take $Q=\mathbb{Z}_N \equiv \mathbb{Z}/(N \mathbb{Z})$, for a genus-g 2-dimensional Riemann surface $\Sigma$, I should have $$H_1(\Sigma; \mathbb{Z}_N)=\prod^{2g}_1 \mathbb{Z}_N,$$ So, $$|H_1(\Sigma; \mathbb{Z}_N)|=|\prod^{2g}_1 \mathbb{Z}_N|=N^{2g}.$$
Question a: more generally, for a generic compact manifold with boundaries: $M$, does this formula work: $$H_1({M},Z_k) =\prod_{k=1}^{b_1({M})} Z_k?$$ so $$|H_1({M},Z_k)| =k^{b_1({M})}?$$
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Question b: Are there extra terms that are missing in the formula above? Can I intuitive see $H_1(\Sigma; \mathbb{Z}_N)$ and $H_1({M},Z_k)$ for a genus-g 2-dimensional Riemann surface $\Sigma$ and for a generic compact manifold $M$ with boundaries in terms of the number of independent cycles of winding around non-contactable loops?
The answer to question a is not always, and the answer to question b is that there may be missing terms.
What you want is the universal coefficients theorem for homology. The general statement is that for any dimension $n$ (in your case $n=1$) and any coefficient group $Q$ (in your case $Q = \mathbb{Z}/N\mathbb{Z}$), and any topological space $\Sigma$ there is a short exact sequence, here we have: $$0 \to H_n(\Sigma) \otimes Q \to H_n(\Sigma;Q) \to \mathrm{Tor}(H_{n-1}(\Sigma);Q) \to 0 $$ The coefficients in "$H_n(\Sigma)$" are, of course, $\mathbb{Z}$. And "Tor" denotes the torsion product operation.
In your case where $n=1$ we have $H_0(\Sigma) \approx \mathbb{Z}$ (assuming path connected for simplicity) and the torsion product term is trivial, so there are no missing terms from that source. But it is possible that $H_1(\Sigma)$ can have some torsion in which case the tensor product $H_1(\Sigma) \otimes Q$ may have missing terms. In particular the projective plane is a counterexample with $Q = \mathbb{Z}/2\mathbb{Z}$, because $b_1=0$.
But if $H_1(\Sigma)$ is torsion free then it works as with an oriented surface (boundary is irrelevant), namely, here we have: $$H_1(\Sigma;Q) \approx H_1(\Sigma) \otimes Q \approx \prod_1^{b_1} Q $$