For a coupling $\pi\in \Pi(\mu,\nu)$ on the product space $(X\times X, \mathcal{F}\times\mathcal{F})$, let $\pi_x$ be the disintegration of $\pi$ with respect to the $\mu$, i.e. there exists a Borel measurable function $x\mapsto\pi_x$ such that $$\pi(dx,dy)=\mu(dx)K(x,dy)=\mu(dx)\pi_x(dy)$$ where $K(\cdot,\cdot)$ is a probability transition kernel.
A question about the notaiton: Why $K$ is called a transition kernel from $(X,\mathcal{F})$ to $(X,\mathcal{F})$, rather than a function $K: X\times X\to [0,\infty)$?
Question: How to get the following inequality:
$$\int W_1(\pi_x, \nu)\mu(dx)\le \int\int d(y,z)\pi_x(dy)\nu(dz)\mu(dx)\le \int d(y,z)\nu(dy)\nu(dz).$$ Why the quality holds as $\pi_x=\delta_{f(x)}$?
Here we replace the coupling $\pi$ by $\pi_x\times \nu$. It seems that we need to show that $$\int \pi_x(dy)\mu(dx)=\int K(x, dy)\mu(dx)=1?$$
Note that $K$ is a probability measure in the second argument, so it is defined on $X\times \mathcal F$ rather than $X\times X$. Precisely, $K(x,A)$ is a (regular) conditional probability that the second coordinate of a vector $(X_1,X_2)$ distributed according to $\pi$ belongs to $A$, given that the first coordinate is equal to $x$. One might think this dynamically: $X_t$ is the state of some system at $t=1,2$. Then $K(x,A)$ is the transition probability for this system from the state $x$ to the set $A$.
The first inequality you are interested in follows from the fact that $\pi_x(dy)\nu(dz)$ is some coupling of $\pi_x$ and $\nu$, so $$W_1(\pi_x,\nu)\le \iint_{X^2} d(y,z)\pi_x(dy)\nu(dz).\tag{1}$$ The second inequality is in fact equality since $\int_{X} \pi_x(dy) \mu(dx) = \nu(dy)$.
For $\pi_x = \delta_{f(x)}$ the inequality in question is equality, since (1) is equality: in this case $\pi_x(dy) \nu(dz)$ is the only coupling of $\pi_x$ and $\nu$.