You can make the connection between classical and quantum field theories through the functional methods. In QFT you have a generating functionals,
\begin{equation}
Z[J]=\int \mathcal{D}\Phi e^{iS[\phi]+iJ\cdot\phi},\quad E[J]=i\ln{Z[J]},\quad J\cdot\phi=\int d^4x\,J(x)\phi(x)
\end{equation}
with $E[J]$ generating connected correlators,
\begin{equation}
(i)^{n+1}\langle\phi(x_1)\ldots\phi(x_n)\rangle_{con,J}=\frac{\delta^n}{\delta J(x_1)\ldots \delta J(x_n)}E[J]
\end{equation}
Let's introduce $\phi_{cl}(x)\equiv\langle\phi(x)\rangle_J=-\frac{\delta E[J]}{\delta J(x)}$. Then we can perform the Legendre transform from function depending on $J$ to function depending on $\phi_{cl}$ introduce the quantum effective action,
\begin{equation}
\Gamma[\phi_{cl}]\equiv -E[J]-J\cdot\phi
\end{equation}
To go back you can always use $\frac{\delta \Gamma[\phi_{cl}]}{\delta\phi_{cl}(x)}=-J(x)$. The important thing for us is that at tree level,
\begin{equation}
\Gamma[\phi_{cl}]\simeq S[\phi_{cl}]
\end{equation}
From here we will make the connection we are looking for.
Start with the classical action $S[\phi]$ and supplement it with some external sources,
\begin{equation}
S_{full}[\phi,J]=S[\phi]+J\cdot\phi
\end{equation}
The classical analog of $E[J]$ is given simply by the value of this full action on the solution of the classical equations of motion $\phi_J$,
\begin{equation}
\frac{\delta S[\phi]}{\delta\phi(x)}\Bigg\vert_{\phi=\phi_J}=-J(x),\quad E_{cl}[J]\equiv-S_{full}[\phi_J,J]
\end{equation}
The variation of this functional will be a classical analogue of the connected correlators. You may ask yourself what it's good for. You may try to find the approximate solution of $\phi_J$ as a series in powers of $J$,
\begin{equation}
\phi_J(x_1)=\phi_0(x_1)+\sum_{n=1}^{+\infty}\frac{1}{n!}\int d^4x_2\ldots d^4x_n\,G_n(x_1,x_2\ldots x_n)J(x_2)\ldots J(x_n)
\end{equation}
As usual in the Taylor series,
\begin{equation}
G_n(x_1,x_2\ldots x_n)=\frac{\delta^{n-1}\phi(x_1)}{\delta J(x_2)\ldots\delta J(x_n)}\Bigg\vert_{J=0}
\end{equation}
On the other hand if you take the variation of $E_{cl}$ you will have,
\begin{aligned}
&\frac{\delta^n E_{cl}[J]}{\delta J(x_1)\ldots \delta J(x_n)}\Bigg\vert_{J=0}=\\
&-\frac{\delta^{n-1}}{\delta J(x_2)\ldots \delta J(x_n)}\left\{\int d^4y\frac{\delta S[\phi]}{\delta \phi(y)}\frac{\delta\phi_J(y)}{\delta J(x_1)}+\int d^4y\, J(y)\frac{\delta\phi_J(y)}{\delta J(x_1)}+\phi_J(x_1)\right\}\Bigg\vert_{J=0}
\end{aligned}
using the definition of $\phi_J$ two first terms cancel each other and we get,
\begin{equation}
\frac{\delta^n E_{cl}[J]}{\delta J(x_1)\ldots \delta J(x_n)}\Bigg\vert_{J=0}=-\frac{\delta^{n-1}\phi_J(x_1)}{\delta J(x_2)\ldots\delta J(x_n)}\Bigg\vert_{J=0}=-G_n(x_1,x_2\ldots x_n)
\end{equation}
I.e. those functions contracted with $J$ will appear in the decomposition of $\phi_J$. Even more usefully if you consider $\phi^n$ theory,
\begin{equation}
-\partial_\mu\partial^\mu\phi(x)-m^2\phi(x)=\frac{g}{(n-1)!}\phi^{n-1}(x)-J(x)
\end{equation}
then in the perturbation theory in $g$ the power of $g$ would also uniquely determine the power of $J$. Thus you may see that the $n$-th power in $g$ will be described by a specific tree level $N$-point connected correlation function which is mentioned in the answers on the question you've linked.
