Let me try and follow the path of the book I mentioned in the comments. This is not an answer (but was to long to fit in a comment) but hopefully it will shed some light on the problem.
Let $(\Omega,\mathcal{F},P)$ be a probability space and $(M,\mathcal{B})$ a measurable space. Let $T:\Omega\to M$ be ($\mathcal{F},\mathcal{B})$-measurable and let $X\in L(P)$. Then we define the conditional expectation of $X$ given $T=t$ to be that/those measurable function(s) $\varphi: M\to\mathbb{R}$ satisfying $ \int_B\varphi(t)\, P_T(\mathrm dt)=\int_{T^{-1}(B)}X\,\mathrm dP,\quad B\in\mathcal{B}.\qquad (*) $ We will write $E[X\mid T=t]:=\varphi(t)$ for any measurable solution to $(*)$. Then $E[X\mid T]=\varphi(T)$ if $\varphi(t)=E[X\mid T=t]$ for all $t\in M$ and $\varphi(t)$ is unique for $P_T$-a.a. $t$.
Let us turn to regular conditional distributions. Let $(L,\mathcal{A})$ be another measurable space and $S:\Omega\to L$ be $(\mathcal{F},\mathcal{A})$-measurable. A regular conditional distribution of $S$ given $T$ is a Markov kernel $P_S^T$ on $(L,\mathcal{A}\mid M,\mathcal{B})$ such that $P_S^T(A\mid t)$ is a conditional distribution of $S$ given $T$, i.e.
1) $Q(\cdot\mid t)$ is a probability measure on $(L,\mathcal{A})$ for all $t\in T$,
2) $Q(A\mid \cdot)$ is $\mathcal{B}$-measurable for all $A\in\mathcal{A}$,
3) $P(S\in A,T\in B)=\int_B P_S^T(A\mid t)\,P_T(\mathrm dt)$ for all $A\in\mathcal{A}$ and $B\in\mathcal{B}$.
If $\psi:L\times M\to \mathbb{R}$ is $(\mathcal{F},\mathcal{A}\otimes\mathcal{B})$-measurable such that $\psi(S,T)\in L(P)$, then
$E[\psi(S,T)]=\int_M \int_L \psi(s,t)\;P_S^T(\mathrm ds\mid t)\, P_T(\mathrm dt),$
$E[\psi(S,T)\mid T=t]=\int_L\psi(s,t)\, P_S^T(\mathrm ds\mid t),$
$E[\psi(S,T)\mid T=t]=E[\psi(S,t)\mid T=t]$
provided that in the equality we use the regular conditional distribution $P_S^T$ to compute the condition expectations.
Isn't it exactly the last equality we are looking for?