Am I being precise to claim the following group isomorphisms: $${SO(3) \times SO(3)}{}=\frac{SO(4)}{\mathbb{Z}_2}$$ $$\frac{SU(2) \times SU(2)}{\mathbb{Z}_2 \times \mathbb{Z}_2 }=\frac{SO(4)}{\mathbb{Z}_2}$$ $$\frac{SU(2) \times SU(2)}{\mathbb{Z}_2}={SO(4)}$$
What are the precise intuitions for the exact isomorphism?
p.s. It seems that some of the previous posts are not very intuitive nor precise:1. recovering-the-two-su2-matrices-from-so4-matrix and 2. Why is SO(3)×SO(3) isomorphic to SO(4)?
The classification of connected Lie groups goes like this: every connected Lie group $G$ has a Lie algebra $\mathfrak{g}$ which determines the isomorphism class of its universal cover $\widetilde{G}$. There is a covering map
$$1 \to \pi_1(G) \to \widetilde{G} \to G \to 1$$
exhibiting $G$ as a quotient of $\widetilde{G}$ by the fundamental group $\pi_1(G)$, which turns out to be a discrete subgroup of the center $Z(\widetilde{G})$ of $\widetilde{G}$. Thus $G$ is classified by the pair consisting of the simply connected Lie group $\widetilde{G}$, together with a particular choice of discrete subgroup $\pi_1(G)$ of the center $Z(\widetilde{G})$.
In this case, all of the Lie groups you describe have the same universal cover, namely $Spin(4) \cong SU(2) \times SU(2)$. The center of this group is $\mathbb{Z}_2 \times \mathbb{Z}_2$ (one $\mathbb{Z}_2$ sitting in each copy of $SU(2)$) and so we can classify all Lie groups with this universal cover by classifying all subgroups of this center (which are all discrete). They are as follows:
So, everything you wrote down is basically fine except that in the third line it's ambiguous what you mean by $\mathbb{Z}_2$. The $\mathbb{Z}_2$ you want is the diagonal one, not the "left" or "right" ones above.