We claim that the only $(n,k)$ that works is $\boxed{(p^2,p+1)}$. Given a duo of unit squares in different columns, there exists exactly one $S_i$ containing both of them. Therefore by double counting these squares, we get $$p^2 \cdot \binom{k}{2} = n \cdot \binom{k}{2}$$which gives $n=p^2$. Further, we can count triples $(x,S_i,S_j)$ such that $x \in S_i \cap S_j$. Choosing $S_i$ and $S_j$ first gives $\binom{n}{2}$ choices. Now, if we choose $x$ first, then for any $y$ that is in a different column from $x$, there is a unique $S_i$ containing both $x$ and $y$. Hence there are $p$ choices for $S_i$ (for a fixed $x$), so $pk \cdot \binom{p}{2}$ for the triple. Equating both gives $k=p+1$. Now for the construction: Label the columns $0$ to $p$ and rows $0$ to $p-1$. For $a,b \in \{0,1, \dots, p-1\}$, let $$S_{(a,b)}=\{(x,ax+b \pmod p) \mid x=0,1, \dots, p-1\} \cup \{(a,p)\}$$These $p^2$ sets satisfy the given conditions because given any two points, there exists a line passing through them, and given a point and a slope, there exists a line passing through that point with a given slope. Further, given two lines, they intersect at exactly one point if they have different slopes, else their corresponding sets intersect in the last column.

Pretty much European Mathematical Cup 2016 J4 and anything classical related to the theory of designs. Nevertheless, nice problem, in case no one has seen it before.