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?
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
$$\int_\mathbb{R} g(y)\,d\mu_2(y)=\int_{\mathbb{R}\times\mathbb{R}}g(y) \, d\mu(x,y) = \int_\mathbb{R} \left(\int_{\mathbb{R}} g(y) \, d(Y_x)_\ast(\mathbb{P})(y)\right)\, dX_\ast(\mathbb{P})(x).$$