Let us consider a system of two spinors. The 3D rotation operator around the $\vec{n}$ axis in $\mathbb{C}^2$ is clearly $R(\theta) = \exp(i \frac{\theta}{2}\vec{n}\cdot\vec{\sigma})$.
If I wish to rotate my system of two spinors around the $z$ axis for instance, which one is the generator of rotations, $S^1_z \otimes S^2_z$ or $S^1_z \otimes \mathbb{I}+ \mathbb{I} \otimes S^2_z$?
It's the second one. To derive this, consider a finite rotation $|\psi_i \rangle \to |\psi' \rangle = U_{ij}(\theta) |\psi_j\rangle$. Acting on a tensor product, we have $$|\psi_i \rangle |\phi_k \rangle \to U_{ij}(\theta) U_{kl}(\theta) |\psi_j\rangle |\phi_l\rangle.$$ If $\theta$ is an infinitesimal angle, we can write $$ U_{ij}(\theta) = \delta_{ij} + \theta S_{ij} + O(\theta^2)$$ where $S$ is a Lie algebra generator. Hence $$|\psi_i \rangle |\phi_k \rangle \to |\psi_i \rangle |\phi_k \rangle + \theta \left[ (S_{ij} |\psi_j \rangle) \otimes |\phi_k \rangle + |\psi_i \rangle \otimes (S_{kl}|\phi_l \rangle) \right] + O(\theta^2).$$ The term inside brackets is precisely of the form $S^1 \otimes \mathbb{1} + \mathbb{1} \otimes S^2$.
