Let $(\Omega,\mathscr{F},P)$ be a probability space.
Let $X,Y$ be random variables on $\Omega$.
Then, we say $Z\sim X|Y$ iff (i) $\int_{Y^{-1}(A)} X dP = \int_{Y^{-1}(A)} Z dP$ and (ii) $Z$ is $\sigma(Y)$-measurable.
Now, let $S:\mathscr{\mathbb{R}} \times \Omega\rightarrow \mathbb{R}$ be a stochastic process.
What does it mean by $X|S$?
There are numerous papers saying like "... because $X|S \sim S$, $P(X\in A|S)= S(A)...$.
I think this is NOT actually a conditional expectation, but it is just a way to denote De Finneti theorem. Isn't it?
Note that $S$ can be seen as a measurable map $\Omega\rightarrow \prod_{A\in \mathscr{B}_{\mathbb{R}}} \mathbb{R}$. If the definition $X|S$ is consistent with the standard conditional expectation definition, $X|S$ is a random variable taking values in $\mathbb{R}$, same as $X$. However, since $X|S\sim S$, $X|S$ must take values in $\prod_{A\in \mathscr{B}_{\mathbb{R}}} \mathbb{R}$. Do you see inconsistency here?
This makes me confusing, so I am curious what's the definition of $X|S$.
What does $X|S$ mean?
For what you have written as $Z \sim X|Y$ the usual notation is $Z=E(X|Y)$. (I am surprised to see your notation).
We can define $E(X|\mathcal G)$ for any sigma field $\mathcal G$ contained in $\mathcal F$. It is defined as a random variable $Z$ measurable w.r.t. $\mathcal G$ such that $\int_A XdP =\int_A ZdP$ for every set $A \in \mathcal G$.
Any stochastic process $S$ gives rise to a sigma algebra $\mathcal G$: the smallest one which makes each random variable in the process measurable. This defines $E(X|S)$