Q1: To get the splitting, you need maps $\mathcal O_S \to j_* \mathcal O_C$ and $j_* \mathcal O_C \to \mathcal O_S$ whose composition $\mathcal O_S \to j_* \mathcal O_C \to \mathcal O_S$ is the identity.
For a map $\mathcal O_S \to j_* \mathcal O_C$, we do something purely formal: A section of $\mathcal O_S$ gives a section of $\mathcal O_C$ on the inverse image, which is the same as a section of $j_* \mathcal O_C$ by definition. This can also be viewed as arising from an adjunction.
For a map $j_* \mathcal O_C \to \mathcal O_S$, we take the trace: $j_* \mathcal O_C $ is a rank two vector bundle on $S$, so every endomorphism of it has a trace in $\mathcal O_S$. But since it has an algebra structure, every section gives an endomorphism.
Examining the composition of these two maps, we see that it is multiplication by two, since each section of $\mathcal O_S$ acts as a diagonal matrix on $j_* \mathcal O_C$, whose trace is the sum of the two identical diagonal entries. So to get a splitting we must divide one of them by two.
Q2: I think it just means nonvanishing away from the branch divisor, with zeroes on the branch divisor.