I’m struggling to understand a step in the application of the following formula to a specific case.
THEOREM. If $$\binom{n}{k}p^\tbinom{k}{2}+\binom{n}{t}(1-p)^\tbinom{t}{2}<1$$ for some $p \in[0,1],$ then $R(k, t)>n$
Let us examine the asymptotics of $R(4, t) .$ We want $\binom{n}{4}p^{6}=c n^{4} p^{6}<1$ so we take $p=\varepsilon n^{-2 / 3}$. Now we estimate $\binom{n}{t}$ by $n^{t}(!), 1-p$ by $e^{-p}$ and $\binom{t}{2}$ by $t^{2} / 2,$ so we want $n^{t} e^{-p t^{2} / 2}<1 .$ Taking $t$-th roots and $\operatorname{logs}, p t / 2>\ln n$ and $t>(2 / p) \ln n=$ $K n^{2 / 3} \ln n .$ Expressing $n$ in terms of $t$ $$ R(4, t)>k t^{3 / 2} / \ln ^{3 / 2} t=t^{3 / 2+o(1)} $$
What happens at that part “expressing $n$ in terms of $t$”?
Before that part, we've used the theorem you've quoted to show that $R(4,t) > n$ provided that $t > K n^{2/3} \ln n$.
We want to show that there is a $k$ such that $n < k (\frac{t}{\ln t})^{3/2}$, then $t > K n^{2/3} \ln n$.
Well, if $n < k (\frac{t}{\ln t})^{3/2}$, then (for $t$ large enough, specifically $t \ge e^{k^{2/3}}$) we also have $n < t^{3/2}$, so $\frac23\ln n < \ln t$. This gives us $$ \frac23n^{2/3}\ln n < k^{2/3} \left(\frac{t}{\ln t}\right) \ln t = k^{2/3}t \implies \frac{2}{3k^{2/3}} n^{2/3}\ln n < t $$ and we get the inequality we want when we take $k = (3K/2)^{-3/2}$, which gives $\frac{2}{3k^{2/3}} = K$.
We conclude that $R(4,t) > n$ for every $n$ such that $n < k(\frac{t}{\ln t})^{3/2}$; in other words, $R(2,t) \ge k(\frac{t}{\ln t})^{3/2}$.