I will give a hand-wavy general argument for the contraposition of the first validity: $\langle R^\ast\rangle \phi \vee \langle S^\ast \rangle \phi \rightarrow \langle (R \vee S)^\ast \rangle \phi$. The main thing to notice is that the set of $R-S$-paths $(R\vee S)^\ast$ includes the sets of $R$-paths and $S$-paths. So, if some $\phi$-state is reachable via an $R$-path, then it is trivially reachable via an $R-S$-path. The same holds for a $\phi$-state reachable via an $S$-path. In general, $\langle R^\ast\rangle \phi \rightarrow \langle (R \vee S)^\ast \rangle \phi$ is valid.
Now, let us consider the contraposition of the second formula: $\langle (R \vee S)^\ast \rangle \phi \rightarrow \langle R^\ast \rangle \phi \vee \langle S^\ast \rangle \phi$. In order to demonstrate that this formula is invalid, it is enough to notice that even though there is an $R-S$-path that reaches a $\varphi$-state, it may be neither $R$- nor $S$-path (i.e. the path alternate $R$ and $S$). A simple counterexample would be a three-state model, where a $p$-state is reachable via two steps, one $R$- and one $S$-step, and the initial and the intermediate states have $\neg p$.