The two random variables give you a probability measure $\mu$ on $\mathcal{B}(S)\otimes\mathcal{B}(T)$. It is enough to get a kernel $\kappa:S\times\mathcal{B}(T)\to[0,1]$ that reproduces $\mu$ when applied to the marginal of $\mu$ on $\mathcal{B}(S)$. Such a kernel is known as a product regular conditional probability.
A sufficient, and potentially necessary, condition is that $\mathcal{B}(S)$ is countably generated and the marginal on that space perfect in the sense of Gnedenko and Kolmogorov, or, equivalently (for e.g. $\sigma$-algebras), compact in the sense of Marczewski. This is shown in The Existence of Regular Conditional Probabilities: Necessary and Sufficient Conditions by Arnold Faden (1979).
A condition that is sufficient and necessary for countably generated probability spaces to admit only perfect probability measures is that $\mathcal{B}(S)$ is universally measurable. A topological condition that guarantees that is being a Hausdorff space that is the image of Baire space $\mathbb{N}^\mathbb{N}$ under a continuous function. A more restrictive condition, but probably the most popular, is that $S$ is Polish, that is separable and completely metrizable.
There is a weaker notion of conditional probability in which the kernel has only to be measurable with respect to the completion of the marginal on $\mathcal{B}(S)$. In this case, one can be much more general and work with spaces that are not countably generated. The seminal paper for this is Existence of Conditional Probabilities by Hoffman-Jorgensen (1971).
Following the notation of the Klenke's book:
Given $B\in \mathcal{E},$ let $\varphi\left(\cdot,B\right)$ be a measurable function such that $\kappa_{Y,\sigma\left(X\right)}\left(\omega,B\right)=\varphi\left(X\left(\omega\right),B\right)$ for all $\omega \in \Omega.$ Note that $\kappa_{Y,\sigma\left(X\right)}\left(\omega,B\right)$ is constant on the set $\{ \omega: X\left(\omega\right)=x\}$ if $x\in \mathcal{E'}$, hence $\varphi\left(x,\cdot\right)$ is a probability measure on $\left(E,\mathcal{E}\right)$ for every $x\in X\left(\Omega\right).$ For every $x\in \left(X\left(\Omega\right)\right)^{c},$ set $\varphi\left(x,\cdot\right)=\mu\left(\cdot\right),$ where $\mu$ is an arbitrary probability measure on $\left(E,\mathcal{E}\right).$ Therefore, $\varphi$ is a stochastic kernel from $\left(E',\mathcal{E}'\right)$ to $\left(E,\mathcal{E}\right).$
By the definition of $\kappa_{Y,\sigma\left(X\right)}$ and the transformation theorem (theorem 4.10):
$$ \mathbb{P}\left(\{Y\in B\}\cap A\right)=\int_{A} \kappa_{Y,\sigma\left(X\right)}\left(\cdot,B\right)d\mathbb{P}= \int_{A} \varphi\left(X\left(\cdot\right),B\right)d\mathbb{P}=\int_{X^{-1}\left(A\right)} \varphi\left(\cdot,B\right)d\mathbb{P}^{X},$$ for every $A\in \sigma\left(X\right).$ Also, by definition:
$$ \mathbb{P}\left(\{Y\in B\}\cap A\right)=\int_{A} \mathbb{P}\left(Y\in B| \sigma\left(X\right)\right)\left(\cdot\right)d\mathbb{P}= \int_{A} \mathbb{P}\left(Y\in B|X=x\right)_{x=X\left(\cdot\right)}d\mathbb{P}=\int_{X^{-1}\left(A\right)}\mathbb{P}\left(Y\in B|X=\cdot\right)d\mathbb{P}^{X}.$$
It is now clear that $\kappa_{Y,X}\left(x,B\right):=\varphi\left(x,B\right)=\kappa_{Y,\sigma\left(X\right)}\left(X^{-1}\left(x\right),B\right)=\mathbb{P}\left(Y\in B|X=x\right)$ for almost $\mathbb{P}^{X}\mbox{-a.a } x\in \mathbb{R}.$
Best Answer
Some thoughts: I'm think it is true when $X$ and $Y$ take values in a Polish space $E$ (or in different Polish spaces). Basically, the conditional independence assumption means that for $A, B \subset E$, we have $\kappa_{\mathcal{F}, (X, Y)}(\omega, A \times B) = \kappa_{\mathcal{F}, X}(\omega, A)\kappa_{\mathcal{F}, Y}(\omega, B)$ for $\omega \in \Omega_{A, B}$, where $P(\Omega_{A, B}) = 1$. But what we want is to find one $\Omega_{0}$ that works for all $A, B$. Taking $\Omega_0 = \bigcap_{A, B \subset E}\Omega_{A, B}$ fails because this is an uncountable intersection. Here I think you can prove $\Omega_0$ exists using second countability, and the $\pi$-$\lambda$ theorem. My thought is that you take a countable base $\{A_1, A_2, \dots\}$ for the topology of $E$ and let $\Omega_0 = \cap_{i, j \geq 1}\Omega_{A_i, A_j}$, but I'll have to work this out some time.