Let $(\Omega, \mathscr{F}, P)$ be a probability space and $\mathscr{G}$ a sub-$\sigma$-algebra of $\mathscr{F}$. If $X$ and $Y$ are (integrable) random variables with $X$ being $\mathscr{G}$-measurable, the pull-out property $\mathbb{E}[XY \mid \mathscr{G}] = X \mathbb{E} [ Y \mid \mathscr{G} ]$ a.s. is well known. Written pointwise, the pull-out property reads as $\mathbb{E}[XY \mid \mathscr{G}](\omega) = X(\omega) \mathbb{E}[Y \mid \mathscr{G}](\omega)$ for almost all $\omega$.
Now assume that $X$ and $Y$ are random elements in some measurable spaces and $X$ is $\mathscr{G}$-measurable. It then follows that $$\mathbb{E} [s(X)r(Y) \mid \mathscr{G} ](\omega) = s(X(\omega)) \mathbb{E} [ r(Y) \mid \mathscr{G} ](\omega) = \mathbb{E} [ s(X(\omega))r(Y) \mid \mathscr{G}](\omega)$$ for almost all $\omega$ where $s$ and $r$ are measurable real-valued functions.
Question: Let $f(x,y)$ be jointly measurable in $x$ and $y$. Is it then true that
$$\mathbb{E}[f(X,Y) \mid \mathscr{G}](\omega) = \mathbb{E}[f(X(\omega),Y) \mid \mathscr{G}](\omega) $$ for almost all $\omega$?
The problem reduces to the question whether the indicator of a measurable set $B$ in the product space can be approximated by linear combinations of indicators over measurable sets of the form $B_1 \times B_2$ or equivalently, whether $f$ can be approximated by finite sums of products $s_k(x)r_k(y)$.
I doubt that this property holds when considering for example $B$ as the diagonal (for $X$ and $Y$ real-valued and not independent).
EDIT:
I think to have found some problems in the statement. The most important problem is that the statement $\mathbb{E} [ f(X,Y) \mid \mathcal{G} ](\omega) = \mathbb{E} [ f(X(\omega), Y) \mid \mathcal{G} ](\omega)$ does not make any sense, since in the expression $f(X(\omega), Y)$ the $\omega$ is not fixed and we can only take conditional expectations over random variables! So, if we consider instead $f(X(\omega_0), Y)$ for some fixed $\omega_0$ then the statement changes to $\mathbb{E} [ f(X,Y) \mid \mathcal{G} ](\omega) = \mathbb{E} [ f(X(\omega_0), Y) \mid \mathcal{G} ](\omega)$ for almost all $\omega$ such that $X(\omega) = X(\omega_0)$, i.e. a.s. on $X^{-1}(\{ X(\omega_0) \} )$. But now another problem appears, since $\{ X(\omega_0) \}$ is in general not contained in the $\sigma$-algebra $\mathscr{E}_X$.
Summing up together: The statement makes sense, if $\mathscr{E}_X$ is atomic. And in this case, the statement is true. The statement does not make sense, if $\mathscr{E}_X$ is not atomic.
EDIT 2:
After some discussions in the comments, it is now clear where the problem is hidden. The expression $\mathbb{E}[f(X(\omega),Y) \mid \mathscr{G}](\omega)$ is to be interpreted as follows. Firstly, for every $x$, define $Z_x(\omega) := \mathbb{E}[f(x,Y) \mid \mathscr{G}](\omega)$ as some fixed(!) chosen version for the conditional expectation of $f(x,Y)$ given $\mathscr{G}$. Note that such a version $Z_x(\omega)$ is chosen for every $x$. Secondly, define $\mathbb{E}[f(X(\omega),Y) \mid \mathscr{G}](\omega) := Z_{X(\omega)}(\omega)$, i.e. by concatenation.
First of all, one has to check whether this map $\omega \mapsto Z(X(\omega),\omega)$ is $\mathscr{G}$-measurable in order to satisfy one of the constraints for the conditional expectation. Assume that it is $\mathscr{G}$-measurable. But then, in general it is NOT a conditional expectation of $\mathbb{E}[f(X,Y) \mid \mathscr{G}]$. The example with $f(x,y) = xy$, $X = Y \sim U(0,1)$ and $\mathscr{G} := \sigma(X)$ in the comments shows that if we choose as versions $Z_x(\omega) := x X(\omega) \mathbf{1}_{X(\omega) \neq x}$ (which depend on $x$!) then the desired equation between the conditional expectations does not hold.
So, this might be the reason, why I have not found this "rule" in mathematical literature on probability theory. However, note that $\mathbb{E}[ f(X,Y) \mid \mathscr{G} ] = \mathbb{E}[ f(x,Y) \mid \mathscr{G}]$ a.s. on $\{ X = x \}$ (which is a direct consequence of the "local property" for conditional expectations). If $\mathscr{G} = \sigma(X)$ then this rule reads as $\mathbb{E} [ f(X,Y) \mid X = x] = \mathbb{E} [ f(x,Y) \mid X = x ]$ and is often referred to as the "replacement rule". But, in order for this rule to be valid, it is necessary that the set $\{ X = x \}$ is measurable (which in general is not true, if $X$ takes values in a general measurable space $(E_X, \mathscr{E}_X)$.