I have posted an identical question in [MSE][1] few days ago, but maybe this site is a better adress to discuss this problem:

Let $G$ be a finite group and $K, H \leq G$ two subgroups. Then
the right quotients $G/H $ and $G/K$ become 
naturally left $G$-sets via $\rho: G \times G/H \to G/H,
(g, fH) \mapsto gfH$  (we will work only with left $G$-sets 
and omit the word left).  
The product $G/H \times G/K$ (or more generally any finite product $\prod_i G/H_i$)
can be naturally endowed
with structure of a $G$-set  by diagonal $G$-action
$(g,(aH, bK)) \mapsto (gaH, gbK)$.

In Tammo tom Dieck's book '[Transformation Groups](https://doi.org/10.1515/9783110858372.312)' it is
stated (Prop 1., page 19) that there are bijections 
between the following sets:

1. $G$-orbits of $G/H \times G/K$.
2. $H$-orbits of $G/K$ with left $H$-action.
3. Double cosets $HgK$, g \in G$.

In the proof the $G$-orbit $G \cdot (eH, gK)$ of $G/H \times G/K$
corresponds to the $H$-orbit $H \cdot gK$ of $G/K$ and to the 
double coset $HgK$.  
This implies that we can decompose $G/H \times G/K$ as set in 
$G$-orbits as follows

$$ G/H \times G/K = 
 	\dot\bigcup_{HgK \in H \backslash G / K}   G \cdot (eH, gK)$$

where $e \in G$ is the identity element and the disjoint 
union runs over all representatives $\overline{g}$ of 
double cosets $HgK \in H \backslash G / K$. 

But we can say even more. Since $G$ acts on any orbit
$G \cdot (eH, gK)$ transitively and the stabilizer
of $(eH,gK)$ is $H \cap gKg^{-1} = H \cap K^{g^{-1}}$, the
orbit $G \cdot (eH, gK)$ is isomorphic to $G/(H \cap K^{g^{-1}})$
and therefore the 
decomposition in $G$-orbits of $G/H \times G/K$ is isomorphic to

$$  \dot\bigcup_{HgK \in H \backslash G / K}   G \cdot (eH, gK) 
G/(H \cap K^{g^{-1}}).$$


**Question:** Is there a natural similar decomposition of the fiber 
product $F:= G/H \times_{G/B} G/K$, which sits in
diagram


$$
\require{AMScd}
\begin{CD}
G/H \times_{G/B} G/K @>{}  >> G/K \\
@VVV  @VVR_y: \ (gK \mapsto gyB)V  \\
G/H @>{R_x: \ (gH \mapsto gxB)}>> G/B
\end{CD}
$$





where the $x, y \in G$ satisfying $x^{-1}Hx \subset B$, $y^{-1}Ky \subset B$ define the $G$-maps 
$R_x: G/H \to G/B$, $gH \to gxB$ and 
$R_y: G/K \to G/B$, $gK \to gyB$.

(Note that up to isomorphism every $G$-map 
$f: G/A \to G/B$ is given by a 
$R_x: G/A \to G/B$, $gA \mapsto gxB$ where $x$ must satisfy
$x^{-1}Ax \subset B$. (Note that $x$ and $x' \in G$ define the same
$R_x$ iff $x'x^{-1} \in B$.))


What we know: set theoretically we have
$F= G/H \times_{G/B} G/K \subset G/H \times G/K$ and 
$F$ consists of pairs $(aH, bK)$ with 
$axB= byB$. Since $R_x$ and $R_y$ are $G$-maps,
if a $(aH,bK) \in F$, then the complete $G$-orbit 
$G \cdot (aH,bK)$ is contained in $F$. Therefore we can
assume $(aH,bK)=(eH, gK) \in F$. Then $xB= gyB$ implies
$x^{-1}gy \in B$. That appears to be our neccessary and
sufficent condition.

So the claim is that $F = G/H \times_{G/B} G/K$ decomposes 
in $G$-orbits as 

$$F = G/H \times_{G/B} G/K = 
\dot\bigcup_{HgK \in H \backslash G / K, \ x^{-1}gy \in B}   
G/(H \cap K^{g^{-1}}).$$

Problem: the parametrizing set 
$HgK \in H \backslash G / K, \ x^{-1}gy \in B$
looks not really nice, especially I would like to get 
rid of the $x^{-1}gy \in B$ condition. Can the fiber product $F = G/H \times_{G/B} G/K$ be 
decomposed in orbits over a more amenable parametrizing set
without this $x^{-1}gy \in B$ condition?

Maybe there exist a parametrization over double coset $H^x \backslash B / K^y$? Can the later be somehow 
related to the double coset $\{HgK \in H \backslash G / K
\ \vert \ x^{-1}gy \in B \}$?





  [1]: https://math.stackexchange.com/questions/4342084/decomposition-of-fiber-product-of-g-sets-in-g-orbits