(I don't speak German, so I don't know what the paper you linked to is about. The following is mostly about the topology of your example, which I'll call $X$.)
Consider a chain of closed subgroups $H\subseteq K\subseteq G$. This gives rise to a so called homogeneous fiber bundle $K/H\rightarrow G/H\rightarrow G/K.$
The projection map sends $gH$ to $gK$. Further, in the case that $H$ is normal in $K$, this fiber bundle is in fact a principal $K/H$ bundle. The element $kH$ acts on $gH$ via $kH *gH = gk^{-1}H$.
This is well defined: if k = k'h_1 and g = g'h_2, then kH * gH = gk^{-1}H = g'h_2h_1^{-1}k'^{-1}H = g'h_3 k'^{-1} H = g'k'^{-1} h_3' H = g'k'^{-1}H. The element $h_3$ is simply, by definition, $h_2h_1^{-1}$ and the element h_3' is defined by k'h_3k'^{-1} = h_3', using normality of $H$ in $K$.
It is also an action as kk'H*gH = gk'^{-1}k^{-1}H = kH*(k'H* gH). Finally, it's free since if k'H gH = gH, then k'^{-1}\in H to k'\in H so k'H = H.
Applying this to your example with $H = SU(p)\times SU(q)$, $K = S(U(p)\times U(q)$, and $G = SU(p+q)$, and noting that $K/H = S^1$ (as Lie groups) via the map sending an element of $S(U(p)\times U(q))$ to the determinant of the $U(p)$ part, we see that $X$ is a principal $S^1$ bundle over the Grassmanian, $Gr$.
In fact, more can be said. Principal $S^1$ bundles are classified by homotopy classes of maps from the Grassmanian into $BS^1 = \mathbb{C}P^\infty = K(\mathbb{Z},1)$. So, canonically, these are classified by $H^2(Gr)$.
It's not too hard to show from the LES in homotopy groups coming from the fibration $S(U(p)\times U(q))\rightarrow SU(p+q)\rightarrow Gr$ that $Gr$ is simply connected with $\pi_2(Gr) = \mathbb{Z}$. It follows from the Hurewicz theorem that $H^2(Gr) = \mathbb{Z}$. (There is also a well known cellular decomposition which gives you this).
Using a similar LES in homotopy groups for $X$, you can see that $H^2(X)$ is trivial. (Further, you can easily that $pi_k(Gr)$ is isomoprhic to $\pi_k(X)$ for all $k\geq 3$.)
It follows that the Euler class of the the principal $S^1$ bundle is a generator of $H^2(Gr)$. This, paired with the Gysin sequence, should allow you to compute the cohomology of $X$ in terms of that of $Gr$.
Finally, if you want to know the characteristic classes of (the tangent bundle to) $X$, notice that $TX = \nu\oplus\pi^* TGr$ where $\nu$ is the trivial rank 1 bundle (coming from the $S^1$ orbits) and that, after using the Gysin sequence, you know what $\pi^*$ does on cohomology.