The map $h: S⁷ \to HP¹ $ given by $(a,b,c,d) \to \frac{a}{c}$ is a fiber bundle with fiber $S^3$ over basepoint $\infty=\frac{a}{0}$. Thus, $h$ induced a sequece $$S^3 \cdots S^7 \to HP¹$$ Even more, this sequence induced a homotopy exact sequence. I would like to construct a similar map between $Sp(2)$ and $HP^1$ with fiber $Sp(1)\times Sp(1)$ and find relationship its with the sequence $$SO(4)\cdots SO(5) \to S^4$$
Are there many references on this subject? I would like to understand the action of these groups and the isomorphism between them.
In general $Sp_n$ acts on $\mathbb{H}P^{n-1}$ in the (almost) obvious way. It acts smoothly and transitively, and the stabiliser of the basepoint $[1,0,\dots,0]\in\mathbb{H}P^{n-1}$ is easily seen to be $Sp_1\times Sp_{n-1}$. Thus, by standard theory there is a smooth, locally trivial fibre bundle $$Sp_1\times Sp_{n-1}\rightarrow Sp_n\rightarrow \mathbb{H}P^{n-1}.$$ This is the quaternionic case of the corresponding fibre bundles $$O_1\times O_{n-1}\rightarrow O_n\rightarrow \mathbb{R}P^{n-1}\qquad U_1\times U_{n-1}\rightarrow U_n\rightarrow \mathbb{C}P^{n-1}.$$ Recall that the projective spaces are Grassmannians of one-dimensional subspaces, after all.
Some further details follow.
Recall that quaternions $\mathbb{H}$ are the normed division algebra of real dimension $4$. The unit sphere $S(\mathbb{H})\subseteq\mathbb{H}$ is thus a group and is homeomorphic (diffeomorphic) to $S^3$. The quaternions are associative but not commutative. Hence we agree to turn $\mathbb{H}^n$ into a quaternionic vector space by allowing $\mathbb{H}$ to act coordinatewise from the right.
Then $Sp_n$ is the group of $\mathbb{H}$-linear transformations $\mathbb{H}^n\rightarrow\mathbb{H}^n$ which preserve the standard quaterionic form. We get a representation for $Sp_n$ as a group of quaternionic matrices by letting these matrices act on $\mathbb{H}^n$ from the left. This was the reason for the bracketed 'almost' in the opening sentence. Basically, because the quaternions are not commutative, you have to pay more attention to whether you take actions from the left of right, and whether you form left or right cosets.
Now I've told you how to from quotients, recall that quaternionic $(n-1)$-space is formed as the right factor space $$\mathbb{H}P^{n-1}=S(\mathbb{H}^n)/S(\mathbb{H})$$ where $S(\mathbb{H}^n)\cong S^{4n-1}$ is the unit sphere in $\mathbb{H}^n$. Since the (left) action of $Sp_n$ on $\mathbb{H}^n$ preserves the unit sphere, and commutes with the right action of $S(\mathbb{H})$, there is a now obvious action on $\mathbb{H}P^{n-1}$. It's easy to see now that the stabiliser over the basepoint is as claimed, and we do indeed get the claimed fibre bundle. Notice that $Sp_1\cong S(\mathbb{H})\cong S^3$.
Now, as a special case we have $$\mathbb{H}P^1\cong S^4$$ This is the same sort of isomorphism as $$\mathbb{R}P^1\cong S^1,\qquad \mathbb{C}P^1\cong S^2.$$ Notice that $S(\mathbb{H})\rightarrow S(\mathbb{H}^2)\rightarrow \mathbb{H}P^1$ is exactly the Hopf fibration $S^3\rightarrow S^7\rightarrow S^4$. We also have the fibration sequence $$Sp_1\times Sp_1\rightarrow Sp_2\rightarrow \mathbb{H}P^1$$ which you were looking for.
What is it's relation to the other fibration sequence you mention? Well, there are exceptional isomorphisms $$Sp_2\cong Spin_5,\qquad Sp_1\times Sp_1\cong S^3\times S^3\cong Spin_4$$ so we can replace the spaces in the last fibration sequence to get something looking like $$Spin_4\rightarrow Spin_5\rightarrow S^4.$$ Normally Clifford theory is used to construct the spinor groups, and the exceptional isomorphisms I mention above come exactly from the relationship between the quaternions and Clifford algebras. What I'm saying, is that with a bit of work you can show that the maps in the last fibration are exactly what you would expect, and it identifies with the standard fibring of $Spin_5$ over $S^4$. Modding out by the central $\mathbb{Z}_2$ action we get $$SO_4\rightarrow SO_5\rightarrow S^4.$$