This is not true without some extra assumptions even in the simplest case where $l=2$ and $t=1$. In this case you count pairs $(k_1,k_2)\in[0,k-1]$ with $(A+k_1)\cap(A+k_2)\ne\varnothing$; that is, with $k_1-k_2\in A-A$. Consider the situation where $n:=|A|\approx 2k^{1/2}$. If $A$ is a block of $n$ consecutive integers, then for $k_1-k_2\in A-A$ to hold you need $|k_1-k_2|<n$, and there are fewer than $2nk\approx 4k^{3/2}$ pairs like this. At the same time, if instead you choose $A$ to have $[-k,k]\subseteq A-A$ (which is possible in view of the assumption $n:=|A|\approx 2k^{1/2}$), then all $k^2$ possible pairs $(k_1,k_2)$ will satisfy $k_1-k_2\in A-A$.

This is not true without some extra assumptions even in the simplest case where $l=2$ and $t=1$. 

In this case you count pairs $(k_1,k_2)\in[0,k-1]^2$ with $(A+k_1)\cap(A+k_2)\ne\varnothing$; that is, with $k_1-k_2\in A-A$. Consider, for instance, the situation where $n:=|A|\approx 2k^{1/2}$. If $A$ is a block of $n$ consecutive integers, then for $k_1-k_2\in A-A$ to hold you need $|k_1-k_2|<n$, and there are fewer than $2nk\approx 4k^{3/2}$ pairs $(k_1,k_2)$ with this property. At the same time, if instead you choose $A$ to have $[-k,k]\subseteq A-A$ (which is possible in view of the assumption $n:=|A|\approx 2k^{1/2}$), then all $k^2$ possible pairs $(k_1,k_2)$ will satisfy $k_1-k_2\in A-A$.