If $\prod_{i=1}^\infty X_i$ is endowed with the product topology, then it suffices to consider open subsets of the form $$U = \left(\prod_{i \le N} U_i \times \prod_{i > N} X_i\right) \times U'$$ $$V = \left(\prod_{i \le N} V_i \times \prod_{i > N} X_i\right) \times V'$$ where $U_i$ and $V_i$ (resp. $U'$ and $V'$) are open subsets of $X_i$ (resp. $X$). As $(\prod_{i=1}^N S_i) \times f$ is topologically transitive (which can be proven by induction using the result you cite in the question since every $S_i$ is topologically mixing and $f$ is topologically transitive), it follows that $f^n(U) \cap V \ne \emptyset$ for some $n \in \mathbf{N}$.

If $\mathcal{X} = (\prod_{i=1}^\infty X_i ) \times X$ is endowed with the product topology, then to show that $F = (\prod_{i=1}^N S_i) \times f$ is topologically transitive (that is, given any pair of open subsets $U, V \subset \mathcal{X}$, there exists $n\in \mathbf{N}$ such that $F^n(U) \cap V \ne \emptyset$), it suffices to consider open subsets of the form $$U = \left(\prod_{i \le N} U_i \times \prod_{i > N} X_i\right) \times U'$$ $$V = \left(\prod_{i \le N} V_i \times \prod_{i > N} X_i\right) \times V'$$ where $U_i$ and $V_i$ (resp. $U'$ and $V'$) are open subsets of $X_i$ (resp. $X$). As $(\prod_{i=1}^N S_i) \times f$ is topologically transitive (which can be proven by induction using the result you cite in the question since every $S_i$ is topologically mixing and $f$ is topologically transitive), it follows that $F^n(U) \cap V \ne \emptyset$ for some $n \in \mathbf{N}$.