$\newcommand{\de}{\delta}$We have
\begin{equation*}
W(\pi):=\frac{N(\pi)}{D(\pi)}, \tag{1}\label{1}
\end{equation*}
where
\begin{equation*}
N(\pi):=\int W_1(\pi_{x_1},\nu)\mu(dx_1)=\int W_1(\pi_{x_1},\pi_2)\pi_1(dx_1),
\end{equation*}
\begin{equation*}
D(\pi):=\int d(y,z)\nu(dy)\nu(dz)=\int d(y,z)\pi_2(dy)\pi_2(dz),
\end{equation*}
$\pi_{x_1}$ is the conditional distribution of $\eta$ given $\xi=x_1$ if the joint distribution of a pair $(\xi,\eta)$ of random variables is the measure $\pi$ with marginals $\mu=\pi_1$ and $\nu=\pi_2$.
The standard notation $\de_a$ is for the Dirac delta measure supported on the singleton set $\{a\}$. So,
\begin{equation*}
\hat{\pi}:=\frac1N\,\sum_{i=1}^N \de_{(\phi(X_i),\,\phi(Y_i))}
\end{equation*}
is the (random) probability measure such that
\begin{equation*}
\hat{\pi}(A)=\frac1N\,\sum_{i=1}^N 1((\phi(X_i),\,\phi(Y_i))\in A) \\
=\frac1N\,\sum_{i=1}^N |\{i\in\{1,\dots,N\}\colon (\phi(X_i),\,\phi(Y_i))\in A\}|
\end{equation*}
for any Borel set $A\in[0,1]^2$. Here $1(\mathcal A)$ is the indicator of an assertion $\mathcal A$, so that $1(\mathcal A)=1$ if $\mathcal A$ is true and $1(\mathcal A)=0$ if $\mathcal A$ is false.
Then for the first marginal $\hat\pi_1$ of $\hat\pi$ and each $x_1\in X=[0,1]$ we have
\begin{equation*}
\hat\pi_1(\{x_1\})=\frac1N\,\sum_{i=1}^N 1(\phi(X_i)\in \{x_1\}) \\
=\frac1N\,\sum_{i=1}^N 1(X_i\in\phi^{-1}(\{x_1\}))
=\frac1N\,\sum_{i=1}^N 1(X_i\in G)
\end{equation*}
for $G=\phi^{-1}(\{x_1\})$,
and for the conditional distribution $\hat\pi_{x_1}$ of $\eta$ given $\xi=x_1$ if the joint distribution of a pair $(\xi,\eta)$ is $\hat\pi$ we have
\begin{equation*}
\hat\pi_{x_1}(B)=\frac{\hat\pi(\{x_1\}\times B)}{\hat\pi(\{x_1\}\times X)}, \tag{2}\label{2}
\end{equation*}
where $B$ is any Borel subset of $X=[0,1]$; if the denominator of the ratio in \eqref{2} is $0$, let $\hat\pi_{x_1}$ be an arbitrary probability measure. Next,
\begin{equation*}
\hat\pi(\{x_1\}\times B)=\frac1N\,\sum_{i=1}^N 1(\phi(X_i)=x_1,\,\phi(Y_i)\in B) \\
=\frac1N\,\sum_{i=1}^N 1(X_i\in\phi^{-1}(\{x_1\}),\,\phi(Y_i)\in B).
\end{equation*}
So, for $G=\phi^{-1}(\{x_1\})$,
\begin{equation*}
\hat\pi_{x_1}(B)=\hat\pi_G(B):=\frac{\frac1N\,\sum_{i=1}^N 1(X_i\in G,\,\phi(Y_i)\in B)}
{\frac1N\,\sum_{i=1}^N 1(X_i\in G)}. \tag{3}\label{3}
\end{equation*}
So,
\begin{equation*}
\begin{aligned}
N(\hat\pi)&=\int W_1(\hat\pi_{x_1},\hat\pi_2)\hat\pi_1(dx_1) \\
&=\sum_{x_1} W_1(\hat\pi_{x_1},\hat\pi_2)\hat\pi_1(\{x_1\}) \\
& =\sum_{G\in \Phi} W_1(\hat\pi_G,\hat\pi_2)\frac1N\,\sum_{i=1}^N 1(X_i\in G) \\
& =\sum_{G\in \Phi} \frac{1}{N}|i\in\{1,\dots,N\}\ \text{s.t.}\ X_i\in G|\,
W_1(\hat{\pi}_G,\hat{\pi}_2).
\end{aligned}
\end{equation*}
Also, if $d$ is the standard metric over $[0,1]$, then, as you have it in your "new edition",
\begin{equation}
\begin{aligned}
D(\hat\pi)&=\int|y-z|\hat\pi_2(dz)\hat\pi_2(dz) \\
&=\sum_{y,z}|y-z|\hat\pi_2(\{y\})\hat\pi_2(\{z\}) \\
&=\sum_{y,z}|y-z|\frac{1}{N^2}\sum_{n,m=1}^N 1(\phi(Y_n)=y,\phi(Y_m)=z) \\
&=\frac{1}{N^2}\sum_{n,m=1}^N \sum_{y,z}|y-z|1(\phi(Y_n)=y,\phi(Y_m)=z) \\
&=\frac{1}{N^2}\sum_{n,m=1}^N |\phi(Y_n)-\phi(Y_m)|.
\end{aligned}
\end{equation}
So,
\begin{equation*}
W(\hat\pi):=\frac{N(\hat\pi)}{D(\hat\pi)}
=\frac{\sum_{G\in \Phi} \frac{1}{N}|i\in\{1,\dots,N\}\ \text{s.t.}\ X_i\in G|\, W(\hat{\pi}_G, \hat{\pi}_2)}{\frac{1}{N^2}\sum_{n,m=1}^N|\phi(Y_n)-\phi(Y_m)|}
\end{equation*}
is a plug-in estimator of $W(\pi)$.
