Why are you ruling out $[2n+1]$? I will change the notation to $K(N,k,s)$ meaning vertices the $\binom{N}{k}$ $k$-subsets of $[N]$ with two connected if they intersect in $s$ or fewer elements.
For a lower bound on the independence number, fix $s+1$ points and take the $\binom{N-s-1}{k-s-1}$ $k$-sets containing those points. This will be the maximum for $N$ sufficiently large with respect to $k$ and , $s.$
Proof Sketch: The bound above is a polynomial of degree $k-s-1$ in $N$. I will show that any independent set not of this type has size less than $2^{3k}\binom{N}{k-s-2}.$ This (outrageous overestimate) is a polynomial (in $N$) of degree less than $k-s-1$ so , for large enough $N$, it is less than $\binom{N-s-1}{k-s-1}.$
Consider an independent set $I$ which includes three $k$-sets (vertices) $A,B,C$ with $|A \cap B \cap C|\leq s.$ Then $I$ consists of $A,B,C$ along with some subset of the set $T$ of vertices which intersect each of $A,B,C$ in $s+1$ or more elements.
Any $D \in T$ includes at least $s+2$ elements of $A \cup B \cup C$ and at most $k-s-2$ elements not in this union. Since $|A \cup B \cup C|<3k.$ There are less than $2^{3k}$ ways to pick a subset of this union. Then there are less than $\binom{N}{k-s-2}$ ways to pick $k-s-2$ more elements. This counts some $D \in T$ multiple times, and it also counts $A,B,C$ and many things not in $T$ (sets which have size less or more than $k$ or do have size $k$ but are not in $T$). In addition, however large $T$ actually is, an independent subset would be quite a bit smaller But still, this count is dominated by $ \binom{N-s-1}{k-s-1}.$
For a specific $k,s$ we could give more sensitive estimates. As an example, consider $k=5,s=1.$
- Fixing two points gives an independent set of size $\binom{N-2}{3}=\frac{N^3}{6}-\frac{3N^2}{2}+\frac{13N}3-4$.
- If we take a fixed set of size $5$ along with any others intersecting it in $3$ or $4$ points, we get an independent set of size $\binom{5}{3}\binom{N-5}{2}+\binom54\binom{N-5}{1}+1=5N^2-50N+126$
For $N<29$ the second scheme is better than the first. After that, the cubic overtakes the quadratic.
For $N=8$ we can take all $\binom{8}{5}=56$ subsets.