random probability measures Consider the discrete random variables $Y,X$ with supports $\mathcal{Y}, \mathcal{X}$. 
Suppose that the probability measure of $X$ conditional on $Y=y$ is itself a discrete random variable with support $\mathcal{A}_{y}$ for any $y\in \mathcal{Y}$ (random probability measure). 
Let $P_{X|y}(x)$ be the probability measure of $X$ conditional on $Y=y$ evaluated at $X=x$.
Let $P_{Y, P_{X|y}}(y_1,a)$ be the joint probability measure of $Y,P_{X|y}$ evaluated at $Y=y_1$, $P_{X|y}(x)=a$.
Questions: 
(1) Should $P_{X|y}(x)$, $X$ and $Y$ be random variables defined on two different probability spaces (one for $P_{X|y}(x)$ and one for $X$ and $Y$)? Could you explain what happens if we use the same probability space $(\Omega, \mathbb{P}, \mathcal{F})$ to define all the three random variables? 
What I am confused about is the following: let $X,Y$ be defined on the probability space $(\Omega, \mathbb{P}, \mathcal{F})$. 
$X$ induces the probability measure $P_X$ such that $P_X(B)=\mathbb{P}(\omega \in \Omega \text{ s.t. } X(\omega)\in B)$ $\forall B\in \mathcal{B}$ where $\mathcal{B}$ is the Borel sigma algebra
Similarly, $Y$ induces the probability measure $P_Y$ such that $P_Y(B)=\mathbb{P}(\omega \in \Omega \text{ s.t. } Y(\omega)\in B)$ $\forall B\in \mathcal{B}$
Lastly, $P_{X|y}(B)=\mathbb{P}(\omega \in \Omega_y \text{ s.t. } X(\omega)\in B)$ $\forall B\in \mathcal{B}$, where $\Omega_y:=\{\omega \in \Omega \text{ s.t. } Y(\omega)=y\}$. 
How can $P_{X|y}(B)$ be defined on $(\Omega, \mathbb{P}, \mathcal{F})$ too?
(2) Is $P_{Y, P_{X|y}}(y_1,a)=0$ by definition?
 A: If a random measure is to be thought of as the conditional distribution of a random variable $X$, then in a way $X$ is a 'random' random variable, and it's hard to separate the different layers of randomness when we talk about $X$.
In your context we are discussing the joint distribution of $Y$ and $P_{X|y}$. This means that necessarily these two objects are defined on the same probability space. As for $X$, notice that $X$ is not explicitly defined here except in terms of its distribution. For example consider a discrete probability space $(\Omega, {\mathbb P}, {\cal F})$ with three objects:
$$
\begin{array}{cccl}
\omega&{\mathbb P}(\omega)&Y(\omega)&P_{X|y}(\omega)\\
\hline
\omega_1&1/4&0&\text{Bernoulli}(p=0)\\
\omega_2&1/4&0&\text{Bernoulli}(p=1)\\
\omega_3&1/2&1&\text{Bernoulli}(p=1/2)\\
\end{array}
$$
This specification is enough to determine $P_{X|y}(x)$, which is a random variable defined by
$$P_{X|y}(x)(\omega):=P_{X|y}(\omega)(\{x\});$$
to emphasize that $P_{X|y}(x)$ is a random variable for each $x$, we could fill out the above table with two additional columns $P_{X|y}(0)(\omega)$ and $P_{X|y}(1)(\omega)$. Similarly the random measure $P_{X|y}(\cdot)$ is defined by
$$
P_{X|y}(B)(\omega):=\sum_{x\in B}P_{X|y}(x)(\omega);
$$
to answer your first question, that's how we define $P_{X|y}(B)$; note that $P_{X|y}(B)$ is a random variable for each $B$.
To answer your second question, no, $P_{Y, P_{X|y}}(y_1,a)$ is not zero by definition. In the discrete probability space defined above, $Y$ has two possible values (namely $0$ and $1$), and $P_{X|y}$ has three possible values (namely, three flavors of the Bernoulli distribution). The joint probabilities of all possible combinations of these will be $0$, or $1/4$, or $1/2$, depending on how you match up $y_1$ and $a$.
