The formula $$r(z) \sim \int_0^{N_S(z)} N_T(z-N_S^{-1}(x)) dx$$ can be re-written in a more appealing way. With the change of variable $u=N_S^{-1}(x)$ it becomes $$r(z) \sim \int_0^{z} N_T(z-u)N'_S(u) du,$$ where $N'_S(u)$ is the derivative of $N_S(u)$ with respect to $u$. With an additional change of variable $u=zv$ it becomes $$r(z) \sim z\int_0^{1} N_T(z(1-v))N'_S(v) dv.$$ Likewise $$t(z) \sim r'(z) = \frac{dr(z)}{dz} =z\int_0^{1} N'_T(z(1-v))N'_S(v) dv .$$
An interesting case is when $S=T$ and $$N_S(u) \sim \frac{a u^b}{(\log u)^c}, \mbox{ with } 0<a, 0<b<1, \mbox{ and } c>0.$$ This covers sums of two primes ($a=1, b=1, c=1$) and sums of two squares ($a=1, b=\frac{1}{2}, c=0$). we have: $$r(z) = \frac{a^2b z^{2b}}{(\log z)^{2c}}\cdot \int_0^1 (1-v)^b v^{b-1}dv = \frac{a^2b z^{2b}}{(\log z)^{2c}}\cdot \frac{\Gamma(b)\Gamma(b+1)}{\Gamma(2b+1)}$$
$$r'(z) = \frac{2 a^2 b^2 z^{2b-1}}{(\log z)^{2c}}\cdot \int_0^1 (1-v)^b v^{b-1}dv = \frac{2a^2 b^2 z^{2b-1}}{(\log z)^{2c}}\cdot \frac{\Gamma(b)\Gamma(b+1)}{\Gamma(2b+1)}$$
Notes
Solutions such as $z=x+y$ and $z=y+x$ count as two solutions: $(x,y)$ and $(y, x)$.
The asymptotic formula for $t(z) \sim r'(z)$, representing the number of solutions to $z=x+y$ with $x\in S, y\in T$ is true only on average, as $z$ becomes larger and larger. There may still be infinitely many integer $z$'s for which $t(z)=0$ even if $r'(z) \rightarrow\infty$ as $z\rightarrow\infty$.
We assume that the sets $S$ and $T$ are "well balanced", for for small and large values of $z$. For instance, if you remove the first $10^{5000}$ elements of $S$, the asymptotic formula for $N_S(u)$ remained unchanged, but this is likely to cause many formulas to fail.
On some tests, I noticed that there are more solutions (on average) to $z=x+y$ with $x\in S, y\in T$ (here $x, y, z$ are integers), if $z$ is even.
If $S=T$ is the set of primes, some adjustments must be made because the primes are not "well balanced", they are less random when they seem (for instance the sum of two odd primes can not be an odd number, but there are also more subtle issues). This is best described in the Wikipedia entry about Goldbach's conjecture (see section about heuristics).
To generate a set like $S$, one way is as follows. Use a random number generator function $U$ returning a uniform deviate on $[0, 1]$. If $U(k) < N'_S(k)$ then add the integer $k$ to the set $S$, otherwise discard it. Do that for all integers.