Since \begin{align} &\operatorname{li}(n)\sim\Pi (n)\equiv\sum _{k=1}^{\lfloor \log (n)\rfloor } \frac{\pi \left(n^{1/k}\right)}{k}\\ \end{align} then the average gap for \begin{align} &\Pi^{-1} (n)\sim\left(\frac{\partial \ \operatorname{li}(n)}{\partial n}\right)^{-1}=\log(n)\\ \end{align}
so, taking \begin{align} &\Pi (n)-\pi(n)=O\left(\frac{\sqrt{n}}{\log n}\right)\\ \end{align} (Ingham), we have \begin{align} &p_n-\Pi^{-1} (n)\sim\left(\frac{\partial }{\partial n}\frac{\sqrt{n}}{\log n}\right)^{-1}=\frac{2 \sqrt{n} \log^2 (n)}{\log (n)-2}\\ \end{align} which gives \begin{align} &g_n\sim\log n+O\left(\sqrt{\log n} \log \log n\right)\\ \end{align} where $g_n$ is the average prime gap, which agrees with Shanks' $\log p(g) \sim \sqrt{g}.$
I don't think the reasoning is correct. It's pretty clear that $$ p_n-\Pi^{-1}(n)\sim\sqrt{\frac{n}{\log n}}\not\sim2\sqrt n\log n\sim\frac{2\sqrt n\log^2n}{\log n-2} $$ and I don't see any proof (or reason to expect proof) that the inverse of the partials gives the average gap size without further assumptions. In any case there's no justification for the last line giving $g_n$.
Instead, note that $(p_2-p_1)+(p_3-p_2)+\cdots+(p_{n+1}-p_n)$ telescopes to $p_{n+1}-p_1$ and so the average of the first $n$ gaps is $$ \frac{p_{n+1}-p_1}{n} $$ which is $$ \log n+\log\log n-1+\frac{\log\log n-2}{\log n}-\frac{(\log\log n)^2-6\log\log n+11}{2\log^2n}+O\left(\left(\frac{\log\log n}{\log n}\right)^3\right) $$ by a result of Cipolla (1902).