Disintegration of pushforward of $(Y\circ (\operatorname{id}\times X))_\#\mathsf P$ Motivation. See my answer here.
Throughout, let $(\Omega, \mathcal A, \mathsf P)$ be a fixed probability space. Let $X$ be a real random variable and assume that we have a family of real random variables $(Y_x)_{x\in\mathbb R}$ that is stochastically independent of $X$. We will define $Y(\omega, x)=Y_x(\omega)$ and assume that $Y$ is $\mathcal A\otimes\mathcal B(\mathbb R)-\mathcal B(\mathbb R)$-measurable. I want to prove that we have the disintegration $$(Y\circ (\operatorname{id}\times X))_\#\mathsf P(\mathrm dy)=(\pi_2)_\#\big(X_\#\mathsf P(\mathrm dx)\otimes Y(\cdot, x)_\#\mathsf P(\mathrm dy)\big).$$
Here:

*

*$\operatorname{id\times X}$ denotes the map from $\Omega$ to $\Omega\times\mathbb R$ mapping $\omega$ to $(\omega, X(\omega))$;

*$\#$ denotes push-forward (for measures). In other words: If we have any measure $\mu:\mathcal A\to[0,\infty]$ and any measurable function $f:\Omega\to\mathbf Z$ for any measurable space $\mathbf Z =(\mathfrak Z, \mathcal B)$, then $f_\#\mu$ is the measure on $\mathbf Z$ defined by $f_\#\mu(B)=\mu(f^{-1}(B))$ for all $B\in\mathcal B$.

*$\pi_2:\mathbb R^2\to\mathbb R$ denotes the projection onto the second coordinate, i.e. $\pi_2(x,y)=y$.

Furthermore, the product between probability measure and stochastic kernel on the right-hand-side is defined as in Korollar 14.23 in the book by Achim Klenke on Wahrscheinlichkeitstheorie (2013). This means that, by Definition, $$\nu\overset{\text{Def.}}=X_\#\mathsf P(\mathrm dx)\otimes Y(\cdot, x)_\#\mathsf P(\mathrm dy)$$ is the unique measure satisfying the regularity from the before-mentioned Korollar such that $$\nu(A\times B)=\int_A Y(\cdot, x)_\#\mathsf P(B)\,X_\#\mathsf P(\mathrm dx)$$ for all Borel-measurable $A,B\subset\mathbb R$.
Therefore, $$(\pi_2)_\#\nu(A\times B) = \nu(\mathbb R\times B) = \int_{\mathbb R} Y(\cdot, x)_\#\mathsf P(B)\,X_\#\mathsf P(\mathrm dx).$$
So in other words we want, for every measurable $B,\subset\mathbb R$, that the following equality holds (both sides possibly being equal to $\infty$):
\begin{equation}\tag{*}\label{*}(Y\circ (\operatorname{id}\times X))_\#\mathsf P(B)=\int_{\mathbb R} Y(\cdot, x)_\#\mathsf P(B)\,X_\#\mathsf P(\mathrm dx).\end{equation}

My attempt. I will show this only if $X$ is a simple random variable, i.e. if its image $\operatorname{im}(X)$ is finite. Then I hope that one can go to general $X$ using some kind of approximation argument.
If $\operatorname{im}(X)$ is finite, then, by Definition, $$(Y\circ (\operatorname{id}\times X))_\#\mathsf P(B)=\mathsf P(\{\omega\in\Omega:Y(\omega, X(\omega))\in B\})=\mathsf P\left(\bigcup_{x\in\operatorname{im}(X)} \{\omega\in\Omega : Y(\omega, x)\in B\text{ and }X(\omega)=x\}\right) = \sum_{x\in\operatorname{im}(X)}\mathsf P\left(\{\omega\in\Omega : Y(\omega, x)\in B\text{ and }X(\omega)=x\}\right). $$
By the stochastical independence assumed above, we have $$\sum_{x\in\operatorname{im}(X)}\mathsf P\left(\{\omega\in\Omega : Y(\omega, x)\in B\text{ and }X(\omega)=x\}\right)=\sum_{x\in\operatorname{im}(X)}\mathsf P(Y(\cdot, x)\in B)\mathsf P(X=x).$$
But the last expression equals $$\int_{\mathbb R} \mathsf P(Y(\cdot, x)\in B) \,X_\#\mathsf P(\mathrm dx)$$ which is equal to $$\int_{\mathbb R} Y(\cdot, x)_\#\mathsf P(B)\, X_\#\mathsf P(\mathrm dx).$$ So the proof is done.

Is my approach saveable and is there an easier proof?
 A: This is true by abstract nonsense, and it seems to me completing your argument would prove the corollary (or the theorem associated to it).

Here are some more details; I'll change the notation a little bit. Let $(\Omega,\mathbb{P})$ be a probability space, $X\in L^0(\Omega;\mathbb{R})$, $Y_\bullet:\mathbb{R}\to L^0(\Omega,\mathbb{R})$ be stochastically independent from $X$ and $Y\in L^0(\mathbb{R}\times \Omega,\mathbb{R})$. Denote by $X_\ast(\mathbb{P})$ the pushforward of $\mathbb{P}$ via $X$ to a probability measure on $\mathbb{R}$. Then we have a diagram in the category of probability spaces

Here $(X,\operatorname{id}_\Omega): \omega\mapsto (X(\omega),\omega)$ (which is your $\operatorname{id}\times X$ up to a shuffle of the coordinates of the target; in my opinion since the map is not $\Omega\times \Omega\to\mathbb{R}\times \Omega$ it's better to not use a product notation here),  $(\operatorname{id}_{\mathbb{R}},Y): (x,\omega)\mapsto (x, Y_x(\omega))$, and $\mu$, $\mu_1$ and $\mu_2$ are the pushforward measures, so that by the functoriality of pushforwards,
\begin{align*}
\mu
&=((\operatorname{id}_\mathbb{R},Y)\circ(X,\operatorname{id}_\Omega))_\ast(\mathbb{P})
=(\operatorname{id}_\mathbb{R},Y)_\ast\circ(X,\operatorname{id}_\Omega)_\ast(\mathbb{P}),\,\, \mu_1 = X_\ast(\mathbb{P})\text{ and }\\
\mu_2
&=(\pi_2)_\ast(\mu)= (\pi_2\circ (\operatorname{id}_\mathbb{R},Y)\circ(X,\operatorname{id}_\Omega))_\ast(\mathbb{P}) 
= (Y\circ (X,\operatorname{id}_\Omega))_\ast(\mathbb{P})
= Y_\ast\circ (X,\operatorname{id}_\Omega)_\ast(\mathbb{P}).
\end{align*}
Disintegrating $\mu$ along $\pi_1$ gives, for $f\in L^0(\mathbb{R}\times\mathbb{R};\mathbb{R})$,
$$\int_{\mathbb{R}\times\mathbb{R}}f(x,y)\, d\mu(x,y)= \int_\mathbb{R} \left(\int_{\mathbb{R}} f(x,y) \, d\mu_x(y)\right)\, d\mu_1(x).$$
The corollary you cite in my notation means that the conditional measure $\mu_x$ of $\mu$ along the fiber of $\pi_1$ at $x$ is precisely $(Y_x)_\ast(\mathbb{P})$, so that we have
$$\int_{\mathbb{R}\times\mathbb{R}}f(x,y)\, d\mu(x,y)= \int_\mathbb{R} \left(\int_{\mathbb{R}} f(x,y) \, d(Y_x)_\ast(\mathbb{P})(y)\right)\, dX_\ast(\mathbb{P})(x).$$
Finally if $g\in L^0(\mathbb{R};\mathbb{R})$, then
$$\int_\mathbb{R} g(y)\,d\mu_2(y)=\int_{\mathbb{R}\times\mathbb{R}}g(y) \, d\mu(x,y) = \int_\mathbb{R} \left(\int_{\mathbb{R}} g(y) \, d(Y_x)_\ast(\mathbb{P})(y)\right)\, dX_\ast(\mathbb{P})(x).$$
