The braid group on $ n $ strands, $ B_n $, is the mapping class group of the disk with $ n $ punctures.
Let $ \Sigma_{g,n} $ denote an oriented surface of genus $ g $ with $ n $ punctures. Let $ M_{g,r} $ denote the mapping class of $ \Sigma_{g,n} $.
I think there is a relationship between $ M_{0,r} $, the mapping class group of the sphere with $ n $ punctures, and $ B_n $, the mapping class group of the disk with $ n $ punctures, but I'm not quite sure what it is.
Examples for small $ n $:
$ n=1 $: $ B_1 $ is trivial and $ M_{0,1} $ is the MCG of the plane so it is also trivial
$ n=2 $: $ B_2 \cong \mathbb{Z} $ and $ M_{0,2} $ is the mapping class group of the cylinder so it is also $ \mathbb{Z} $
$ n=3 $: $ B_3 $ is well known to be a the universal central extension $ \mathbb{Z}: PSL(2,\mathbb{Z}) $. According to MCG $ M_{0,3} \cong S_3 $. In general there is canonical map $ B_n \to S_n $. So $ M_{0,3} $ is a quotient of $ B_n $.
$ n=4 $: $ B_4 $ has a $ B_3=\mathbb{Z}: PSL(2,\mathbb{Z}) $ quotient. Since $ M_{0,4} \cong 2^2 : PSL(2,\mathbb{Z}) $ by MCG it is plausible that again $ M_{0,4} $ is a quotient of $ B_4 $.
Is it true in general that $ M_{0,n} $ is a quotient of $ B_n $? Is there a natural way to describe this surjection $ B_n \twoheadrightarrow M_{0,n} $? Is there anything interesting we can say about the kernel?
I'm not sure this will satisfy you, but, I would say that your indexing is off by $1$.
The most special relation that I know of between braid groups and mapping class groups of punctured surfaces is the relation between $M_{0,n}$ and $B_{n-1}$, which goes like this.
Let me model \begin{align*} M_{0,n} &= \mathcal{MCG}_+\bigl(S^2 - \{q_1,...,q_n\}\bigr) \\ B_{n-1} &= \mathcal{MCG}_+\bigl(D^2 - \{p_1,...,p_{n-1}\}, \partial D^2\bigr) \end{align*} where the latter formula represents the group of orientation preserving homeomorphisms of $D^2-\{p_1,...,p_{n-1}\}$ that fix $\partial D^2$ pointwise, modulo isotopies that leave points of $\partial D^2$ stationary.
$M_{0,n}$ has a homomorphism onto the symmetric group $S_n$ obtained by recording how the punctures $\{q_1,...,q_n\}$ are permuted. Let $M^1_{0,n}$ denote the subgroup fixing $q_1$; the index of this subgroup in $M_{0,n}$ is $(n-1)!$.
There is a central extension $$1 \mapsto \mathbb Z \mapsto B_{n-1} \mapsto M^1_{0,1} \mapsto 1 $$ The central subgroup $\text{image}(\mathbb Z \to B_{n-1})$ is generated by the Dehn twist around $\partial D^2$. The quotient homomorphism $B_{n-1} \mapsto M^1_{0,1}$ is induced by any choice of continuous function $$D^2 - \bigl( \{p_1,...,p_{n-1}\} \bigr) \to S^2 - \{q_1,...,q_{n-1}\} $$ that restricts to a constant map $\partial D^2 \mapsto q_n$ and also restricts to a homeomorphism $$D^2 - \bigl(\{p_1,...,p_{n-1} \cup \partial D^2\} \bigr) \mapsto S^1 - \bigl(\{q_1,...,q_{n-1}\} \cup \{q_n\}\bigr) = S^1 - \{q_1,...,q_n\} $$