(ii) $\implies$ (i)
Suppose $\mu = \sum_{n=1}^\infty \mu_n$, where $\mu_i \perp \mu_j$ for all $i \neq j$. By definition of "mutually singular," there exist pairwise disjoint $A_n \in M$ such that $\mu_n(A) = \mu_n(A \cap A_n)$ for all $A \in M$. Therefore, $\mu(A_k) = \sum_{n=1}^\infty \mu_n(A_k) = \mu_k(A_k) < \infty$, so that each $A_n$ has finite $\mu$-measure. Moreover, $\mu(A) = \sum_{n=1}^\infty \mu_n(A_n) = \sum_{n=1}^\infty \mu_n(A \cap A_n) = \sum_{n=1}^\infty \mu(A \cap A_n)$, where in the last step we used the fact that $\mu(A \cap A_n) = \mu_n(A \cap A_n)$ for all $n \in N$.
(i) $\implies$ (ii)
Let $\mu_n$ be a measure concentrated on $A_n \in M$, normalized so that $\mu_n(A_n) = 1$. Then $\mu(A) = \sum_{n=1}^\infty \mu(A\cap A_n) = \sum_{n=1}^\infty \mu_n(A)$.
(EDIT following Nate's comment: We can define $\mu_n(A) = \mu(A \cap A_n)$.)