How do we gain an intuitive visualization that the rotational symmetry group of $S^3$ as $$SO(4)=(S^3\times S^3)/\mathbf{Z}/2?$$
techinically I know that there are following isomorphism and diffeomrphisms of these differentiable manifolds:
$$Spin(4)=SU(2) \times SU(2)=(S^3\times S^3).$$ $$SO(4)=Spin(4)/\mathbf{Z}/2=(SU(2) \times SU(2))/\mathbf{Z}/2=(S^3\times S^3)/\mathbf{Z}/2.$$ $$PSO(4)=Spin(4)/(\mathbf{Z}/2)^2=(SU(2) \times SU(2))/(\mathbf{Z}/2)^2=(SO(3)\times SO(3)).$$
How do we show the above can all be intuitive visualized?
This isomorphism is known as the isoclinic decomposition; the wikipedia page on $SO(4)$ describes it in more detail. While it may or may not count as an "intuitive visualization", it can be described quite compactly using quaternions:
We can identify $\mathbb{R}^4$ with the space of quaternions $\mathbb{H}$, and $S^3$ with the group of unit quaternions. $S^3$ acts on $\mathbb{H}$ by left multiplication and right multiplication, giving two actions $R,L$. Since these actions are norm preserving and $S^3$ is connected, both map into $SO(4)$. Further, since the actions commute, they give rise to an action $\rho:S^3\times S^3\to SO(4)$ defined by $\rho_{(a,b)}q=L_aR_bq=aqb^{-1}$. One can show that $\rho$ is surjective and $\ker(\rho)=\{(1,1),(-1,-1)\}$.