I think you mean $D\cup\{{0\}}$, not $D\cap\{{0\}}$.
Let's look at an example, $t=2$. Then if I understand your notation, we are working modulo $4t-1=7$, and we have a set with $2t-1=3$ elements, and every non-zero number comes up $t-1=1$ time as a difference. An example would be the set $D=\{{1,2,4\}}$, where we get the differences $1=2-1$, $2=4-2$, $3=4-1$, $4=1-4$, $5=2-4$, and $6=1-2$.
Now let's take the union with zero, $\{{0,1,2,4\}}$. The size of the set is now $2t=4$, and we get the new differences $1=1-0$, $2=2-0$, $3=0-4$, $4=4-0$, $5=0-2$, and $6=0-1$, so now every nonzero number comes up $t=2$ times as a difference.
So the thing you have to understand is why when you look at the numbers $d-0$ and $0-d$ for $d$ in $D$ you get each nonzero number exactly once. That has something to do with the special nature of prime powers congruent to $3$ modulo $4$, and the special nature of the Paley difference sets. So, do you understand how the Paley sets are constructed?