@Megan, I think you meant in your original statement

Statement 1: Let $r$ and $s$ be fixed numbers. Then KST theorem says that for some constant $c = c(r,s)$, a graph with at least $c(r,s)n^{2-\frac{1}{r}}$ edges contains a copy of $K_{r,s}$.

[At any rate, it really should be made clear that the constant $c$ is growing with $s$ and $r$. I found myself confused by the original wording at least.]

If a graph w $c(r,s)n^{2-\frac{1}{r}}$ edges contains a copy of $K_{r,s}$ then for any positive $c'$, a graph in $G$ with $c(r,s)n^{2-\frac{1}{r}} + c'n$ edges definitely contains $c'n$ copies of $K_{r,s}$ given KSK stated above. Indeed, you can find a copy of $K_{r,s}$ in $G$ and remove an edge from the copy, and then find a copy of $K_{r,s}$ in the remaining graph ($G$ minus the edge removed), and so on and so forth, $c'n$ times, to get $c'n$ distinct copies of $K_{r,s}$.

If $c'$ is no larger than 1, then $c'n$ is small relative to $c(r,s)n^{2-\frac{1}{r}}$ so the following can be concluded from KSK theorem as written above plus the above paragraph:

Statement 2: Let $r$ and $s$ be fixed numbers. Then for some constant $\tilde{c} = \tilde{c}(r,s)$, a graph with at least $\tilde{c}(r,s)n^{2-\frac{1}{r}}$ edges contains at least $n$ copies of $K_{r,s}$.

In fact, you can use the above line of reasoning plus Statement 1 to show the existence of as many as $c(r,s)n^{2-\frac{1}{r}}$ copies of $K_{r,s}$ where $c(r,s)$ is as in Statement 1, but if you want that many copies then $\tilde{c}(r,s)$ grows to $2c(r,s)$ as in Statement 1.

the above argument probably can be tightened further.