Suppose you are given two metric spaces $(X,d_{X})$ and $(Y,d_{Y})$ and a map $f \colon X \to Y$. Furthermore take two measures $\mu, \nu$ in $P_{1}(X)$ the Wasserstein 1-space over $X$. Let $\gamma \in \Lambda(\mu , \nu)$ be a coupling. I have seen that $(f\times f)_{*}\gamma$ is a coupling of $f_{*}\mu$ and $f_{*}\nu$, but I am confused about how exactly to write down a proof of this.
I thought about the following. If $A$ is in the Borel $\sigma$-algebra of $X$ we have: $$(f\times f)_{*}\gamma(A\times X) = (f\times f)_{*}\mu(A)=f_{*}\mu (A)$$ and then doing the same with $\nu$, but I am suppose that is not correct since $(f\times f)_{*}\mu$ may not be defined at all.
A second question is about the Wasserstein 1-distance of this push-forward measures.
We have seen in class that $$\int_{Y\times Y} d_{Y}(y_{1},y_{2})\, \text{d}(f\times f)_{*}\gamma)(y_{1},y_{2}) = \int_{X\times X} d_{Y}(f(x_{1}),f(x_{2}))\, \text{d}\gamma(x_{1},x_{2}).$$ I would like to proof this equation but I have no idea which argument I should use.
Considering the first question, note that $f_*\mu$ and $f_*\nu$ are measures on $Y$, so the left hand side of the equation you wrote doesn't make sense. Even if you replaced $A\times X$ with $B\times Y$, where $B\subset Y$, the further equalities don't make sense because, as you correctly pointed out, $(f\times f)_*\mu$ is not defined.
To see why $(f\times f)_*\gamma$ is a coupling of $f_*\mu$ and $f_*\nu$, it suffices to use the definition of push-forward measures: $$ (f\times f)_*\gamma(B\times Y) = \gamma((f\times f)^{-1}(B\times Y)) = \gamma(f^{-1}(B)\times X) = \mu(f^{-1}(B)) = f_*\mu(B), $$ and similarly $(f\times f)_*\gamma(Y\times B) = f_*\nu(B)$.
Considering the second question, it's a general fact that $$ \int_Y g\ d (f_*\mu) = \int_X g\circ f\ d\mu.$$ See Pushforward measure integral property.