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I know that the rotation operator on a single qubit about axis $\hat{n}$ by an angle $\theta$ can be represented as $ \exp(-i \theta \hat{n}\cdot\vec{\sigma}/2)$. With two qubits, what action on Bloch sphere vectors does \begin{equation} \exp(i \theta \hat{n_1} \cdot \vec{\sigma_1} \otimes \hat{n_2}\cdot \vec{\sigma_2}/2) \end{equation} represent?
Certainly not simultaneous rotation of both qubits by $\theta$ (according to some quick and dirty numerics).
If you have two qubits, then the state space does not consist of two separate copies of the Bloch sphere: the Hilbert space is the tensor product of the component two-state spaces, which doesn't admit a clean representation (and certainly not as clean Bloch spheres). If you have local unitaries you can picture them using the local Bloch spheres but that's not the case for your entangling unitary.
Instead, the operation you mention is best thought of as a somewhat modified version of the standard entangling gate CZ (controlled-Z, itself a close relative of the controlled-NOT gate). That is best seen by noting that your operation, which reads $$ \exp(i \theta \sigma_{1z} \sigma_{2z}/2) $$ in a suitable reference frame, can be transformed, using only local operations, to the explicit form $$ \exp(i \theta |{\uparrow} \rangle \langle{\uparrow} | \otimes|{\uparrow} \rangle \langle{\uparrow} |/2) $$ of a (continuous) controlled-phase gate.
