Let $ B_n $ be the braid group on $ n $ strands. Let $ MCG(\Sigma_g) $ be the mapping class group of a surface of genus $ g $.
Is there a relationship between $ B_{2g+1} $ and $ MCG(\Sigma_g) $?
For $ g=1 $ we have $ MCG(\Sigma_1)=MCG(T^2)=SL(2,\mathbb{Z}) $ while $ B_3 \cong \widetilde{SL(2,\mathbb{Z})} $. For $ g=0 $, $ MCG(\Sigma_0)=MCG(S^2) $ and $ B_1 $ are both trivial. (Here we take the convention that the mapping class group is orientation preserving).
What about $ B_5 $ and $ MCG(\Sigma_2) $? Is there some relationship?
Does it have something to do with ramified coverings/Riemann-Hurwitz? Or with the symplectic representation of the braid group Symplectic representation of the braid group?
There is indeed a relation between $B_5$ and $MCG(\Sigma_2)$. I believe you can find the details in Birman's book, and perhaps also in the book of Farb and Margalit.
In brief, $MCG(\Sigma_2)$ has a normal subgroup that is cyclic of order $2$, the nontrivial element of which is referred to as the "hyperelliptic involution". The quotient of $MCG(\Sigma_2)$ by this subgroup is the mapping class group of the 6-punctured sphere, $MCG(S^2 - \{\text{6 points}\})$. On the other hand, $B_5$ has an infinite cyclic center, and the quotient by $B_5$ by its center is isomorphic to a finite index subgroup of $MCG(S^2 - \{\text{6 points}\})$, namely the subgroup which stabilizes one of the 6 deleted points.
Outside of these "low complexity" coincidences, no, I would say there is not any relationship between $B_{2g+1}$ and $MCG(\Sigma_g)$. The reason for the relationship in the case $g=2$ has to do with a very exceptional situation, namely that $MCG(\Sigma_2)$ has an order 2 center. This can be exploited to learn a lot about $MCG(\Sigma_2)$ (see again Birman's book), but no similar exploit is available for $MCG(\Sigma_g)$ when $g \ge 3$.