Q1: To get the splitting, you need maps $\mathcal O_{\mathbb P} \to j_* \mathcal O_C$ and $j_* \mathcal O_C \to \mathcal O_{\mathbb P}$ whose composition $\mathcal O_{\mathbb P} \to j_* \mathcal O_C \to \mathcal O_{\mathbb P}$ is the identity.
For a map $\mathcal O_{\mathbb P} \to j_* \mathcal O_C$, we do something purely formal: A section of $\mathcal O_{\mathbb P}$ 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 $j_* \mathcal O_C \to \mathcal O_{\mathbb P}$, we take the trace: $j_* \mathcal O_C $ is a rank two vector bundle on ${\mathbb P}$, so every endomorphism of it has a trace in $\mathcal O_{\mathbb P}$. 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 $\mathcal O_{\mathbb P}$ 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.