= Solution
The two unheaded requests are addressed here; the individual group-cover calculations are in the scoped parts below. A <quaternion> $q=a+bi+cj+dk$ has <norm> $|q|^2=a^2+b^2+c^2+d^2$. Thus the <unit quaternions> form exactly the sphere $S^3\subset\mathbb R^4$, with <quaternion> multiplication giving its smooth group structure. The explicit identification with the <special unitary group> $SU(2)$ is
$$
q\longmapsto
\begin{pmatrix}
a+ib&c+id\\
-c+id&a-ib
\end{pmatrix}.
$$
The <matrices> representing $i,j,k$ satisfy the <quaternion> multiplication relations. The displayed <matrix> obeys $U^\dagger U=|q|^2I$ and $\det U=|q|^2$; conversely every $SU(2)$ <matrix> has that form with $|q|=1$. Hence \b[$S^3$ is the manifold of <unit quaternions>, and its group is $SU(2)$].
For the <exterior-square double cover of SO(3,3)>, fix a <basis> $e_1,\ldots,e_4$ and volume element $v=e_1\wedge e_2\wedge e_3\wedge e_4$. Define the symmetric bilinear form $Q$ on the six-dimensional exterior square by
$$
\alpha\wedge\beta=Q(\alpha,\beta)v.
$$
It is symmetric because <two-forms> commute under the <wedge product>. Put
$$
a_1=e_1\wedge e_2,\quad a_2=e_1\wedge e_3,\quad a_3=e_1\wedge e_4,\qquad
b_1=e_3\wedge e_4,\quad b_2=-e_2\wedge e_4,\quad b_3=e_2\wedge e_3.
$$
Then $Q(a_i,a_j)=Q(b_i,b_j)=0$ and $Q(a_i,b_j)=\delta_{ij}$. The Gram <matrix> is $\begin{pmatrix}0&I_3\\I_3&0\end{pmatrix}$: the vectors $(a_i+b_i)/\sqrt2$ have <norm> $+1$, while $(a_i-b_i)/\sqrt2$ have <norm> $-1$. Therefore \b[the quadratic form $\omega\mapsto\omega\wedge\omega$ has signature $(3,3)$].
For $A\in SL(4,\mathbb R)$,
$$
(\Lambda^2A\,\alpha)\wedge(\Lambda^2A\,\beta)
=\Lambda^4A(\alpha\wedge\beta)=\alpha\wedge\beta.
$$
Thus $\rho(A)=\Lambda^2A$ preserves $Q$. The <special linear group> is connected: <polar decomposition of an invertible real matrix> writes $A=OP$ with $O\in SO(4)$ and $P$ positive definite of <determinant> one, and both factors can be joined to the identity. The image consequently lies in $SO_0(3,3)$.
If $\Lambda^2A=I$, then $Au\wedge Aw=u\wedge w$ for every pair, so $A$ preserves every two-plane. A line is an intersection of two such planes, hence every line is preserved. A <linear map> preserving every line is scalar: apply it to <basis> vectors and then to their pairwise sums. Thus $A=\lambda I$, and $\Lambda^2A=I$ forces $\lambda^2=1$. The kernel is exactly $\{\pm I\}$. It is discrete, so the derivative of $\rho$ is injective. Both <Lie algebras> have dimension $15$, making the image open; an open subgroup of a connected group is the whole group. Hence
$$
\boxed{SO_0(3,3)\cong SL(4,\mathbb R)/\{\pm I\}.}
$$
The identity-component restriction is essential. The same principle of an explicit form-preserving representation, kernel calculation and dimension argument supplies the five remaining covers.
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