Let $G$ be a group and fix subgroups $G', H, K \leq G$ and additional subgroups $H' \leq H$ and $K' \leq K$, where $H' : = H \cap G'$ and $K' : = K \cap G'$; further suppose that all of these groups are non-trivial. There is a natural map $$ H' \setminus G' / K' \longrightarrow H \setminus G /K $$ between the $(H', K')$ double cosets of $G'$ and $(H , K)$ double cosets of $G$. I would like to show that this map is injective (or at least injective for some specific groups I am working with, but I believe this will work for arbitrary groups).
To this end, let $x , y \in G'$. Since $H' \leq H$, $G' \leq G$, and $K' \leq K$, we certainly have $H' x K' \subseteq HxK$ and $H'yK' \subseteq H y K$. So if it is the case $H x K = H y K$, then $H' x K' , H' y K' \subseteq H x K = H y K$. This does not seem strong enough to conclude that $HxK = HyK$ implies $H'xK' = H'yK'$ because, while $H'xK'$ and $H'yK'$ are both subsets of the same $(H , K)$ double coset of $G$, they could still be disjoint.
Is this claim true?
In general, the map you described is not injective, here is a way to construct some counter-examples:
Let $A$ be an arbitrary group and let $G = A \times A \times A$, $K = A \times \{1\} \times A$, $H = \{1\} \times A \times A$ and $G' = \{(a,a) \:|\: a \in A\} \times A$, then $H' = K' = \{1\} \times \{1\} \times A$ and so $H'\backslash G' / K'$ is in bijection with $A$ whereas $H \backslash G/K$ is a singleton set. Thus, as long as $A$ is not a singleton the map $H'\backslash G' / K' \to H\backslash G / K$ is not injective.