Does the following identity hold? I encountered a claim in the proof a a claim in a paper that seemed to imply/use the following:

Suppose we have two subgroups $K$ and $H$ of a larger group $G$.  Let $T$ be a translation of $K$, which is defined as a subset of $G$ such that every $g \in G$ can be uniquely written as $t k$, for some $t \in T$ and $k \in K$. For example, if $G = \mathbb{Z} / 12 \mathbb{Z}$ and $K = \{0, 2, 4, \cdots, 10\}$, then $T = \{0, 1\}$ is a possible translation of $K$.
Let us also for simplicity define the notation $t K = \{tk ~|~ k \in K\}$.
The claim is then the following: for any translation of $T$ and any subgroup $H \leq G$ which is neither a subset nor a superset of $K$, we have the following:
$$\sum_{t \in T} |tK \cap H|^2 = |H| \cdot |K \cap H|$$

Is this claim true? It does seem to work out for example with $G, T$ as given above and $H = \{0, 3, 6, 9\}$, but I can't seem to find a proof of this fact or even an intuition as to why this should be true. If it is not true, then are there any other conditions we can impose (maybe I have missed them in reading the paper) so that it is true?
EDIT: The paper in question can be found on ArXiv at this link. The question stems from a step in the proof of lemma $4$.
 A: This is correct. In order to prove it, we need a lemma first:
I claim that for any $t\in G$ either $tK\cap H$ is empty or $|tK\cap H|=|K\cap H|$. To prove this, let $|K\cap H|=n$ and then let $k_1,k_2,\dots,k_n$ be the distinct members of $K\cap H$. And assume $tK\cap H$ is nonempty, so there is some $y=tk\in tK\cap H$. Then the elements $yk_1,yk_2,\dots yk_n$ must all be distinct. But $y,k_i\in H$, so $yk_i\in H$, and $yk_i=(tk)k_i=t(kk_i)\in tK$. So we have identified $n$ distinct elements of $tK\cap H$, which means that if we let $m=|tK\cap H|$ then $m\geq n$.
Now, let $tk_1,tk_2,\dots,tk_m$ be the distinct elements of $tk\cap H$. But if $tk_1\in H$ then $(tk_1)^{-1}=k_1^{-1}t^{-1}\in H$, and so the elements $(tk_1)^{-1}tk_1,(tk_1)^{-1}tk_2,\dots,(tk_1)^{-1}tk_m$ must all be distinct elements of $H$. But we can simplify those elements down to $k_1^{-1}k_1,k_1^{-1}k_2,\dots,k_1^{-1}k_m$, which clearly must be elements of $K$. So we have identified $m$ distinct elements of $K\cap H$, meaning that $n\geq m$, and therefore $n=m$.
Now, we prove the main result. We know that the cosets $tK,\ t\in T$ are a partition of $G$, and therefore the sets $tK\cap H$ are a partition of $H$. Let $T'=\{t\in T|tK\cap H\neq \varnothing\}$. Then if $|K\cap H|=n$, we have for every $t\in T'$ that $|tK\cap H|=n$. We also know that $|H|=mn$ for some integer $m$, and therefore $|T'|=m$. Finally, we know that restricting $t$ to $T'$ doesn't change the sum $\sum |tK\cap H|^2$, since if $t\not\in T'$ then $|tK\cap H|=0$. Putting it all together:
\begin{align*}
\sum_{t\in T} |tK\cap H|^2&=\sum_{t\in T'} |tK\cap H|^2\\
&=\sum_{t\in\{t_1,t_2,\dots,t_m\}}n^2\\
&=mn^2\\
&=n\cdot(mn)\\
&=|K\cap H|\cdot |H|
\end{align*}
A: This is basically the same argument as Benjamin's, but I dislike putting group elements in lists.
Let $T'=\{t\in T\mid tK\cap H=\varnothing\}$ as in the other answer.
Not only is $|tK\cap H|$ always either $0$ or $|K\cap H|$, but it's a coset of $K\cap H$. That is, if $s\in tK\cap H$ (i.e. the intersection is nonempty) then $s\in tK$ means $tK=sK$ and $s\in H$ means $H=sH$ so
$$ tK\cap H=sK\cap sH=s(K\cap H). $$
This yields a map $T'\to H/(K\cap H)$. It's actually a one-to-one correspondence.
Since the $tK$s (hence the $tK\cap H$s) are disjoint, this map is one-to-one, and it's also onto since $G/K$ is a partition so every coset of $H/(K\cap H)$ intersects some $tK$ (hence some $tK\cap H$).
This means we have
$$ \sum_{t\in T'} |tK\cap H|^2=\left|\frac{H}{K\cap H}\right||K\cap H|^2 = |H||K\cap H|. $$
