As it happens, $f(30,4)=20.$ It is a fairly unique construction based on some almost magic seeming coincidences. You can check that $f(4,3)=6.$ Given a base set $S=\{a,b,c,d,e,f\}$ there are $\binom63=20$ triples. I'll write $abc$ (in any order) for the triple $\{a,b,c\}.$

Two systems showing $f(4,3)=6$ are $\{ace,adf,bcf,bde\}$ and $\{acf,ade,bce,bdf\}.$ They come in a certain way from the partition $ab|cd|ef$ into disjoint pairs. In all there are $15$ such partitions and $30$ such systems. The details are an easy exercise.

So for $f(30,4)=20$ I take the $20$ "points" to be the triples from $S$ and the $30$ "lines" to be the quadruples of points forming a system as above for $f(4,3)=6.$ Again, this construction does not generalize to other cases.

I use the language above to help point to the literature. I'll be a little loose with the definitions here and let you check out all the conditions. Say that a *design* (another name is *hypergraph*) is a family $\mathcal{B}$ of $b$ subsets called blocks from a base set $V$ with $v$ points. I've replaces $r$ with $b.$ I will also replace $n$ with $k$ and call the design $k$-uniform if all the blocks have size $k.$

A design, which might or might not be uniform, is called a *linear space* if every pair of points belongs to a unique block. In that case the blocks are often called *lines* since two points determine a unique line. A $k$-uniform linear space is called a Balanced Incomplete Block Design BIBD-$(v,b,k,r,1)$ or a Steiner $k$ system. The $r$ is the constant (in this case) number of blocks containing each point. The parameters are related by $vr=bk$ and $v(v-1)=bk(k-1).$ So given $b,k$ both $v$ and $r$ are determined. I think that for fixed $k$ there is such a system provided $b$ is large enough and the values of $v$ and $r$ are integers.

If there is such a system then $f(b,k)=v.$ There can't quite be one with with $b=30,k=4$ since that gives $v(v-1)=360$ but $19\cdot 18=360-19$ and $20\cdot19=360+20.$ However $v=20$ might be just right if we allow each point to have a unique complement not in any block with it. And that works.

A partial linear space (which might be uniform) is like a linear space except that two points determine at most one line. So you have defined $f(b,k)$ to be the smallest $v$ for which there is a $(b,v,k)-$partial linear space. Since we can always throw away some blocks, $f(b-1,k) \le f(b,k)$ so I would expect that $f(b-j,k)=f(b,k)=v$ in the event that there is a BIBD-$(b,v,k,r,1)$ and $j$ is small enough.

The example at the top was a rather nice partially balanced incomplete block design with $\lambda_1+\lambda_2=1.$ That paper shows a few nice constructions. It is from 1955 so more is known now then then, but it is a good place to look.