$\textit{Q1:}$ Why the statement below holds? If someone could give the intuition and/or a proof I would appreciate it. (this one is answered already so check $Q2$)
Suppose that $\mathcal{I}=(X,\mu)$ and $\mathcal{J}=(Y,\nu)$ are measurable probability spaces, where $\mu$ and $\nu$ denote the probability distributions over $X$ and $Y$ respectively. Thus, if $\phi:X\to\Delta(Y)$, where $\Delta(Y)$ is the simplex of $Y$, then if the image of $\mu$ by $\phi$ is $\nu$ then it holds $$\mathbb{E}_{\mu}\phi(x)(y)=\nu(y)$$
Note that $\mu$ and $\nu$ are probability measures s.t. for every $x_i$ such that $\mu(x_i)>0$, then $p(x_i)\in\Delta(X_{-i})$ denotes the conditional probability of $\mu$ given $x_i$ over $X_{-i}$: $$p(x_i)(x_{-i})=\mu(x_{-i}|x_i)=\frac{\mu(x_{-i},x_{i})}{\mu(x_i)}$$
Also, note that $\mu(x_i)$ stands for $\mu(\{x_i\}\times X_{-i})$.
If $i=\{1,2,3\}$, then $X=X_1\times X_2\times X_3= \{a_1,b_1\}\times\{a_2,b_2\}\times\{a_3,b_3\}$ and $x=(\underbrace{a_1}_{x_1},\underbrace{a_2}_{x_2},\underbrace{a_3}_{x_1})$
$\textit{Q2:}$ In the second part I can not clarify from the notation how are the ex-post and ex-ante probabilities connected with $\phi$, $\mu$, $\nu$ and $p$?
Even furhter, the following probabilities are defined. Let $r$ denote the ex-ante probability and $q$ the ex-post probability, they are defined as follows on $\Delta(Y_{-i})$:
$$r(x_i)(y_{−i})= P_{\phi}(y_{−i}|x_i),\quad\text{and}\quad q(y_i)(y_{−i})= P_{\phi}(y_{−i}|y_i)$$
for $P_{\phi}(x_i,y_i)>0$, where $P_{\phi}(x_i,y_i)=\mu(x_i)\phi_{i}(x_i)(y_i)$ then $r(x_i)$ and $q(y_i)$ are random vectors with values in $\Delta(Y_{-i})$ and $f(y_i|x_i)=\phi_{i}(x_i)(y_i)$ and $r(x_i)=\mathbb{E}_{p(x_i)}\phi_{-i}(x_{-i})$
I think that $q(y_i)(y_{−i})=\frac{\nu(y_{-i},y_i)}{\nu(y_i)}$, but what is the formula of $r(x_i)(y_{−i})$ and how it ends up to $r(x_i)=\mathbb{E}_{p(x_i)}\phi_{-i}(x_{-i})$
$\textit{Hint:}$ Note that $\phi(x)(y)$ means that if $x$ is drawn according to $\mu$, then $y$ is drawn according to $\nu$ with probability $\phi(x)(y)$ where $\phi(x)$ denotes the probability distribution according to which $y$ is going to drawn in other words $\phi(x)(y)=\phi(y|x)$. I think that this $\phi$ is called transition probability between the two probability meaures.
As far as I understand it, the equation $\Bbb E_\mu \phi(x)(y)=\nu(y)$ is just a restatement of the law of total probability. To find the probability of $y$ occuring, that is, to find $\nu(y)$, you sum over all of the ways that $y$ can occur. That is, for each $x\in X$, you take the probability of getting $x$, $\mu(x)$, and multiply by the conditional probability of getting $y$ given $x$, which is $\phi(x)(y)$. The result is $$ \nu(y)=\sum_{x\in X}\mu(x) \times \phi(x)(y) $$ But the RHS is of the form $\sum_{x\in X} \mu(x) f(x)$, where $f(x)=\phi(x)(y)$, and this is by definition (or really, by the law of the unconscious statistician), equal to $E_\mu[\phi(x)(y)]$.
In other words, there is nothing really to prove. The equation $\nu(y)=E_\mu \phi(x)(y)$ is just a mathematical translation of the previous statement "The image of $\mu$ by $\phi$ is $\nu$."