easy to check $h_n(G)=\frac{5}{4}+O\left(\frac{1}{n}\right)$ and that $G$ is $H$-free.

Now we prove the upper bound. Consider any $H$-free hypergraph $G$ of edge-type $\{2,3\}$ on $n$ vertices. Recall that $d_{2}(v)$ denotes the number of 2-edges that contain $v$. For each pair of 2-edges that intersect $v$ there is a unique 3-set containing those two pairs. This 3-set cannot appear in the edge set of $G$ since $G$ is $H$-free. We say that the edge is forbidden. Note that each 3-edge may be forbidden up to 3 times in this manner–depending on which of the three vertices we call $v$. Hence there are at least $\frac{1}{3} \sum_{v\in [n]} \binom{d_{2}(v)}{2}$ 3-edges which are not in $G$. Note that this is true for any $H$-free graph $G$ with number of vertices. Hence $$\begin{aligned}
h_n(G) &\leq \frac{ \binom{n}{3}-\frac{1}{3}\sum_{v\in [n]} \binom{d_{2}(v)}{2} }{\binom{n}{3}} + \frac{ \frac{1}{2}\sum_{v\in [n]} d_{2}(v)}{\binom{n}{2}} \\                
&= 1-\frac{\sum_{v\in [n]} d_2(v)^{2}}{6\binom{n}{3}} +\left(\frac{1}{6\binom{n}{3}}+\frac{1}{2\binom{n}{2}}\right)\sum_{v\in [n]} d_2(v) +1.\end{aligned}$$ Applying Cauchy-Schwarz Inequality and letting $m:=\sum_vd_2(v)$, we have $$\begin{aligned}
h_n(G) &\leq \frac{-\left(\sum_{v\in [n]} d_2(v)\right)^2}{6n\binom{n}{3}}
+\left(\frac{1}{6\binom{n}{3}}+\frac{1}{2\binom{n}{2}}\right)\sum_{v\in [n]} d_2(v) + 1 \\
&=-\frac{m^2}{n^4} + \frac{m}{n^2}+1 +O\left(\frac{1}{n}\right)\\
&\leq \frac{5}{4}  +O\left(\frac{1}{n}\right).\end{aligned}$$ In the last step, we use the fact that $f(x)=1+x-x^2$ has the maximum value $\frac{5}{4}$. Taking the limit, we get $\pi(S(K_{2}^{\{1,2\}}))\leq \frac{5}{4}.$

We 