I have been given a function $\varphi(A,B):\mathbb{H}\to\mathbb{H}$, $h\mapsto AhB^{-1}$ where $A,B\in SU(2)\times SU(2)$. I don't understand how this forms a well-defined map, nor how this would even return quaternion.
Going off the answer given here, it is straightforward enough to define map $F(C,D):h\mapsto ChD^{-1}$ given $C,D\in SU(2)$ - this clearly yields a quaternion. However, I can't then assume that $Ah$ is $ChD^{-1}$.
Every quaternion $q \in \mathbb{H}$ can be thought of as a special complex $2$ by $2$ matrix of the form:
$$\left( \begin{array}{cc} u & -\bar{v} \\ v & \bar{u} \end{array} \right) $$
where $q = u + j v$. One can check the algebra of such matrices is isomorphic to $\mathbb{H}$. Note that the group of unit quaternions can be shown to be isomorphic to $SU(2)$.
So essentially the map they are talking about can be understood either in a purely quaternionic way, where you multiply a quaternion from the left and right by unit quaternions, or using complex two by two matrices. I hope I made things a bit clearer.