Now, take two copies of this spherical triangle (corresponding to two copies of $\mathbb CP^2$ - and if we want to be careful I think they have to have opposite orientations), remove a neighborhood of one vertex in each triangle (corresponding to removing a ball in each copy of $\mathbb CP^2$, since the orbits of these vertices are points) and glue them together along this deleted neighborhoods. The resulting Alexandrov space is a rectangle (since the two remaining vertices on each of the triangles glued together had angle $\pi/2$); and this is exactly the orbit space of the cohomogeneity $2$ torus action on $\mathbb C P^2\#\mathbb CP^2$. In [Since you were asking, in particular, you get lots of circle sub actions, if you want, obviously but with larger orbit spacesspaces.]
This picture is similar to the one you had of the cohomogeneity one manifold $S^3\times_{S^1} S^2$ in the following way. Imagine we are foliating the rectangle (orbit space of $T^2$-action on $\mathbb C P^2\#\mathbb CP^2$) by vertical segments, from side to side. Each of the vertical (actually, any) sides of the rectangle is a cohomogeneity one manifold with singular orbits equal to points and principal isotropy group a circle. Indeed, they are equivariantly diffeomorphic to $S^2$ with the $S^1$ rotation action. In this way, you get two $S^2$'s on opposite sides, and in the "middle", we're left with vertical segments that correspond to cohomogeneity one manifolds with singular orbits equal to circles and trivial principal isotropy groups. They are in fact spheres $S^3$ with the standard $T^2$-action. So we have a similar picture: two $S^2$'s on opposite sides and $S^3$'s in the middle. But this is different from the cohomogeneity one manifold $S^3\times_{S^1} S^2$, where you have an $S^3$ action, even though it has a similar topological structure of two $S^2$'s on opposite sides and $S^3$'s filling the "middle". Sorry for the rather informal description, I hope it helps...

