Non-trivial $S^2$-bundle over $S^2$ So every $S^2$-bundle over $S^2$ is either trivial or $\mathbb C P^2\#-\mathbb C P^2$ i.e. $\mathbb CP^2$  blown-up at a point (see this question)
My question; Given the group $G$ with Lie algebra $\mathfrak g=\mathfrak {sl}(2,\mathbb C)\oplus \mathfrak {su}(2)$.
and the bundle $$F\hookrightarrow E \to B$$
where $F:=\mathrm{SL}(2,\mathbb C)\Big/\begin{pmatrix} *&*\\ 0&*\end{pmatrix}\; \cong S^2$
and  $B:=\mathrm{SU}(2)\Big/\begin{pmatrix} *&0\\ 0&*\end{pmatrix}\; \cong S^2$. Can we construct a subgroup $H$ of $G$ such that $G/H\cong \mathbb C P^2\#-\mathbb C P^2$? 
 A: In fact, the classification of compact simply connected homogeneous $4$-manifolds is quite easy to state: there are only three up to diffeomorphism and they are $S^4, \mathbb{C}P^2$, and $S^2\times S^2$.
The proof is not too bad, compared to, say, the $5$-dimensional classification.  (But even that proof is not so bad...)
So, suppose a Lie group $G$ acts transitively on a simply connected closed $4$-manifold $M$.  We will assume throughout that the ineffective kernel $K = \{g\in G: gp = p \text{ for all }p\in M\}$ of the action is at most finite.  This will allow us to pass to and from covers without worrying.
Montgomery proved the following:

Suppose $G$ is a Lie group acting transitively on a simply connected closed manifold $M$.  Then the identity component of $G$ also acts transitively. Further, the maximal compact subgroup of the identity component acts transitively.  Even more so, if we find a cover of $G$ which splits as $G = T^k\times G_0$ with $G_0$ simply connected, then $G_0$ acts transitively.

Thus, we may restrict attention to the case where $G$ is a compact simply connected Lie group.  Compactness allows us to assume the action is isometric by averaging an arbitrary Riemannian metric.  Let $p\in M$ and set $H = \{g\in G: gp = p\}$, the isotropy group of the $G$ action on $M$ at $p$.
The map $H\rightarrow O(T_p M)$ given by $h\mapsto d_p h$ has finite kernel, so up to finite cover, $H$ embeds into $O(T_p M) = O(4)$.
If we follow the proof I gave here, it now follows that the rank of $H$ is at most $2$ and that some cover of $H$ has the form $T^{b_2(M)}\times H_0$ with $H_0$ a simply connected Lie group.  This already shows $b_2(M)\leq 2$, and now the rest of the classification breaks down by cases depending on $b_2(M)$.
Since you care about $\mathbb{C}P^2 \sharp -\mathbb{C}P^2$, let me restrict attention to the case $b_2(M) = 2$ (which happens to be the easiest of the three cases).
Since we know the rank of $H$ is at most $2$ and that, up to cover, $H = T^2 \times H_0$, it follows that $H = T^2$ on the nose.
Because $G/H = M$ is a $4$-manifold, $\dim G = 6$.  Up to cover, there are just not very many $6$-dimensional closed Lie groups, and only one of them is simply connected:  $G$ must be $SU(2)\times SU(2)$.  So, $G$ has rank $2$, so $H$ is a maximal torus.  By the maximal torus theorem, any two maximal tori are conjugate (and it's easy to check that $G/H$ is naturally diffeomorphic to $G/(gHg^{-1})$ for any $g\in G$), so we can replace $H$ by our favorite maximal torus in $G$.
My favorite maximal torus is the product of the usual $S^1\subseteq SU(2)$.  But then $$M = G/H = (SU(2)\times SU(2))/(S^1\times S^1) = (SU(2)/S^1)\times (SU(2)/S^1) = S^2\times S^2.$$
