If a set $S$ is evenly covered by lines in $n$ slopes, then $n \le |S|-1$ because through every point $p$, there are at most $|S|-1$ lines connecting $p$ to other points in the set, and any other slope of line would include a line intersecting $S$ in just $p$.

Here are some examples achieving that bound: Hyperovals in subfields.

A hyperoval in the projective plane over a finite field $\mathbb{F}_q$ of characteristic $2$ is a set of $q+2$ points in the projective plane over $\mathbb{F}_q$ so that every line meets the set in $0$ or $2$ points. An example is $\lbrace(1,t,t^2)~\bigg|~t\in\mathbb{F}_q \rbrace \cup \lbrace(0,1,0)\rbrace \cup \lbrace(0,0,1)\rbrace$. For some $q \gt 8$ there are many non-isomorphic hyperovals.

There are $q+2 \choose 2$ lines out of $q^2+q+1$ that intersect the hyperoval in $2$ points. The remaining $q \choose 2$ lines do not intersect the hyperoval. The complement of a line disjoint from the hyperoval gives a set of $q+2$ points in the affine plane over $\mathbb{F}_q$ which intersects every line in $0$ or $2$ points. For example, if $q=2$, this is the whole plane.

Let $\mathbb{F}_{q'} \subset \mathbb{F}_q$. We can construct a hyperoval within the subplane $\lbrace (a,b)|a,b \in \mathbb{F}_{q'} \rbrace$ consisting of $q'+2$ points so that for every slope in $\mathbb{F}_{q'} \cup \lbrace \infty \rbrace$, the lines of that slope intersect the hyperoval in $0$ or $2$ points. If you don't like the vertical lines of slope $\infty$, then if $q' \lt q$, you can take an automorphism of the plane taking a slope in the complement $\mathbb{F}_q \setminus \mathbb{F}_{q'}$ to $\infty$, and this will take the hyperoval to a set evenly covered by $q'+1$ parallelism classes of finite slopes.

So, if $q=2^k$ then for every $d|k, d\lt k$ there are sets of $2^{d}+2$ points evenly covered by $2^{d}+1$ slopes. This includes $d=1$: $\lbrace (0,0), (1,0),(x,1),(x+1,1) \rbrace$ for some $x \in \mathbb{F}_q \setminus \lbrace 0,1 \rbrace$, which is evenly covered by the slopes $0$, $1/x$, and $1/(x+1)$.