So recast the problem as:

$Q(n,k,m,t):$ Given the family of *all* $k$ element subsets of $\{{1,\cdots,n\}}$, how large can a sub-family $\mathcal{F}$ be if we require that no $m$ of them have an intersection of size $t$?

Then your question is $Q(n,k,k',k')$. A very rough bound is that at the absolute largest we could imagine that every subset of of size $t$ is in $m-1$ members of $F$. If $N$ is the size of $F$ then counting in two ways all ways to select a member of $\mathcal{F}$ along with a further subset of size $t$ we have that $$N\binom{k}{t} \le (m-1)\binom{n}{t}$$ so $$N \le \frac{(m-1)\binom{n}{t}}{\binom{k}{t}} \tag{*}$$

It might be the case that that is not as crude as one might think in the sense that one can prove that there is for fixed $k,m,s$ always a family of size $N$ (rounded down to an integer) provided that $n$ is larger than some $n(k,m,s)$. However that could be a result of the form "there is at least one way to pick the family $\mathcal{F}$ which works" without any real guidance on how to get an actual specimen (nor an assurance that there are any examples with a compact description).

To get equality in $(*)$ requires the right hand side to be an integer (of course). Then $\mathcal{F}$ is called a $t$-design (with $\lambda=m-1$) and a Steiner system when $m=2$. A very exciting recent paper (link and description here) claims that this integrality condition is sufficient for existence, provided that $n$ is large enough.