@Anne has answered my question in a comment. I formalize her ideas as below.
We have $L^\infty(Y)=(L^1(Y))^*$, so
$$
\|\pi_\sharp f\|_{L^\infty(Y)} = \sup \{\langle \pi_\sharp f, g \rangle_{L^1(Y)} ; \|g\|_{L^1(Y)} = 1\}.
$$
We have the duality
$$
\langle \pi_\sharp f, g \rangle_{L^1(Y)} = \langle f, \pi^\sharp g \rangle_{L^1(X)}.
$$
So
$$
\|\pi_\sharp f\|_{L^\infty(Y)} = \sup \{\langle f, \pi^\sharp g \rangle_{L^1(X)} ; \|g\|_{L^1(Y)} = 1\}.
$$
Notice that
$$
\begin{align}
\langle f, \pi^\sharp g \rangle_{L^1(X)} &\le \|f\|_{C(X)} \langle \pmb 1, \pi^\sharp g \rangle_{L^1(X)} \\
&= \|f\|_{C(X)} \int_X \pi^\sharp g \mathrm d \mu \\
&= \|f\|_{C(X)} \int_X g \circ \pi \mathrm d \mu \\
&= \|f\|_{C(X)} \int_Y g \mathrm d \nu \quad (\star)\\
&\le \|f\|_{C(X)} \|g\|_{L^1(Y)}.
\end{align}
$$
Here $\pmb 1$ is the constant map $x \mapsto 1$. Also, $(\star)$ follows from change-of-variables formula. This completes the proof.
Update 1: @Anne has pointed out a mistake in above proof, i.e., the duality holds for $g \in L^2(Y)$ but not necessarily for $g \in L^1(Y)$. However, it seems we can obtain the same inequality with a tweak on the definition of $\pi^\sharp$. We define a map
$$
\pi^\sharp: L_1(Y) \to L_1(X), g \mapsto g \circ \pi.
$$
Then $\pi^\sharp$ is a linear operator. Moreover,
$$
\|\pi^\sharp\| = \sup_{g \in L_1(Y)} \frac{\|g \circ \pi\|_{L_1(X)}}{\|g\|_{L_1(Y)}} = \sup_{g \in L_1(Y)} \frac{\int_X |g| \circ \pi \mathrm d \mu}{\int_Y |g| \mathrm d \nu} = \sup_{g \in L_1(Y)} \frac{\int_Y |g| \mathrm d \nu}{\int_Y |g| \mathrm d \nu} = 1.
$$
It follows that $\pi^\sharp$ is bounded. We denote its adjoint by
$$
\pi_\sharp : L_\infty(X) \to L_\infty(Y).
$$
We have the duality
$$
\int_Y (\pi_\sharp f) g \mathrm d \nu = \int_X f (\pi^\sharp g) \mathrm d \mu \quad \forall f \in L_\infty(X), \forall g \in L_1(Y). \quad (\star)
$$
For $f \in C(X)$, we have $f \in L_\infty(X)$, and
$$
\begin{align}
\|\pi_\sharp f\|_{L_\infty(Y)} &= \sup \left \{ \int_Y (\pi_\sharp f) g \mathrm d \nu \,\middle\vert\, \|g\|_{L_1(Y)} = 1 \right\} \\
&= \sup \left \{ \int_X f(\pi^\sharp g) \mathrm d \mu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \quad \text{by } (\star)\\
&\le \|f\|_\infty \sup \left \{ \int_X \pi^\sharp g \mathrm d \mu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \\
&= \|f\|_\infty \sup \left \{ \int_X g \circ \pi \mathrm d \mu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \\
&= \|f\|_\infty \sup \left \{ \int_Y g \mathrm d \nu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \quad (\star\star)\\
&\le \|f\|_\infty \sup \left \{ \int_Y |g| \mathrm d \nu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \\
&= \|f\|_\infty.
\end{align}
$$
Here $(\star\star)$ follows from change-of-variables formula. This completes the proof.
Update 2: I show here why Tao's function $\pi_\sharp f\in L_2(Y)$ is equal $\nu$-a.e. to my function $\pi'_\sharp f\in L_\infty(Y)$. We have
$$
\int_X f(\pi^\sharp g) \mathrm d \mu = \int_Y\left(\pi_{\sharp} f\right) g \mathrm d \nu \quad \forall g \in L_2 (Y).
$$
$$
\int_X f(\pi^\sharp g) \mathrm d \mu = \int_Y \left(\pi'_{\sharp} f\right) g \mathrm d \nu \quad \forall g \in L_1 (Y).
$$
Because $\nu$ is finite, we get $L_\infty(Y) \subset L_2(Y) \subset L_1(Y)$. So
- $\pi_{\sharp} f - \pi'_{\sharp} f \in L_2(Y)$.
- $$
\int_Y (\pi_{\sharp} f - \pi'_{\sharp} f) g \mathrm d \nu =0 \quad \forall g \in L_2 (Y).
$$
Pick $g:= \pi_{\sharp} f - \pi'_{\sharp} f$, we get
$$
\int_Y (\pi_{\sharp} f - \pi'_{\sharp} f)^2 \mathrm d \nu =0
$$
It follows that $\pi_{\sharp} f - \pi'_{\sharp} f=0$ $\nu$-a.e.
Because $X$ is compact, $C(X)$ is separable. Let $F$ be a countable dense subset of $C(X)$. Let $\mathcal F := \operatorname{span}_{\mathbb Q} (F)$. For each $f \in F$, we have $\pi_\sharp f$ is an equivalence class of $L_\infty (Y)$, so we fix a representative $\tilde \pi_\sharp f \in \mathcal L_\infty (Y)$. For each $y \in Y$, we define a map $L_y:\mathcal F \to \mathbb C$ by
$$
L_y \bigg (\sum_{i=1}^n \lambda_i f_i \bigg) := \sum_{i=1}^n \lambda_i (\tilde \pi_\sharp f_i) (y) \quad \forall (\lambda_i)_{i=1}^n \subset \mathbb Q, \forall (f_i)_{i=1}^n \subset F.
$$
Let $\pmb \lambda = (\lambda_i)_{i=1}^n \subset \mathbb Q$ and $\pmb f=(f_i)_{i=1}^n \subset F$ such that $f := \sum_{i=1}^n \lambda_i f_i \in F$. Because $\pi_\sharp$ is linear,
$$
\tilde\pi_\sharp f = \tilde\pi_\sharp \bigg (\sum_{i=1}^n \lambda_i f_i \bigg) = \sum_{i=1}^n \lambda_i (\tilde\pi_\sharp f_i) \quad \nu \text{-a.e.}
$$
So there is a $\nu$-null set $N_{(\pmb \lambda, \pmb f)} \in \mathcal Y$ such that
$$
\tilde\pi_\sharp f (y) = \sum_{i=1}^n \lambda_i (\tilde\pi_\sharp f_i) (y) \quad \forall y \in Y \setminus N_{(\pmb \lambda, \pmb f)}.
$$
Because the set of such pairs $(\pmb \lambda, \pmb f)$ is countable, there is a $\nu$-null set $N_1 \in \mathcal Y$ such that for each $y\in Y \setminus N_1$, $L_y$ is well-defined and $\mathbb Q$-linear on $\mathcal F$.
Let $\pmb \lambda = (\lambda_i)_{i=1}^n \subset \mathbb Q$, $\pmb f=(f_i)_{i=1}^n \subset F$, and $f:= \sum_{i=1}^n \lambda_i f_i$. Because $\pi_\sharp$ is $\mathbb Q$-linear and $\|\pi_\sharp f\|_{L_\infty(Y)} \le \|f\|_\infty$, we get
$$
\sum_{i=1}^n \lambda_i (\tilde\pi_\sharp f_i) = \tilde\pi_\sharp (f) \le \|f\|_\infty \quad \nu \text{-a.e.}
$$
So there is a $\nu$-null set $N_{(\pmb \lambda, \pmb f)} \in \mathcal Y$ such that
$$
\sum_{i=1}^n \lambda_i (\tilde\pi_\sharp f_i) (y) \le \|f\|_\infty \quad \forall y \in Y \setminus N_{(\pmb \lambda, \pmb f)}.
$$
Because the set of such pairs $(\pmb \lambda, \pmb f)$ is countable, there is a $\nu$-null set $N_2 \in \mathcal Y$ such that for each $y\in Y \setminus N_2$, $L_y (f) \le \|f\|_\infty$ for all $f \in \mathcal F$.
Let $N := N_1 \cup N_2$. Then $N \in \mathcal Y$ is a $\nu$-null set such that for each $y \in N^c := Y \setminus N$, we have
- $L_y$ is well-defined and $\mathbb Q$-linear.
- $L_y$ is bounded, i.e., $L_y (f) \le \|f\|_\infty$ for all $f \in \mathcal F$.
Let $\mathcal G := \operatorname{span}_{\mathbb R} (F)$. Then $\mathcal F$ is dense in $\mathcal G$. So for each $y \in N^c$, we extend $L_y$ to a $\mathbb R$-linear continuous function on $\mathcal G$. In particular, if $(f_n) \subset \mathcal F$ and $f \in \mathcal G$ such that $f_n \to f$, then
$$
L_y (f) := \lim_{n \to \infty} L_y (f_n).
$$
Best Answer
Let $(\mu_y)_{y\in Y}$ be the family obtained by above theorem. Let $G := X \times Y$ and $\mathcal G$ the Borel $\sigma$-algebra generated by the product topology. We define a map $$ \phi : X \to G, x \mapsto (x, \pi (x)). $$
Let $\lambda := \phi_\sharp \mu$ be the push-forward of $\mu$ through $\phi$. Then $$ \lambda (C) = \int_Y (\mu_y \otimes \delta_y) (C) \mathrm d \nu (y) \quad \forall C \in \mathcal G. $$
Notice that $\mathcal G$ coincides with $\mathcal X \otimes \mathcal Y$, so the product measure $\mu_y \otimes \delta_y$ is compatible with $\mathcal G$. Let $K$ be a countable union of compact subsets of $X$ that supports $\mu$; $K$ is Lusin and thus Suslin.
By Suslin-Lusin theorem, $H := \phi (K)$ is Suslin.
Notice that $G$ is separable. By Lemma, $H$ is analytic in $G$. Let $\hat{\mathcal G}$ be the universal completion of $\mathcal G$ and $\hat \lambda$ the unique extension of $\lambda$ from $\mathcal G$ to $\hat{\mathcal G}$. It is mentioned here that measurable sets are analytic, and all analytic sets are universally measurable. So $H\in \hat{\mathcal G}$. Clearly, $\hat \lambda (H) = 1$. Then there is $M \in \mathcal G$ such that $M \subset H$ and $\lambda(M)= 1$. We have $$ 1=\int_Y (\mu_y \otimes \delta_y) (M) \mathrm d \nu (y). $$
It follows that $$ (\mu_y \otimes \delta_y) (M) = 1 \quad \nu\text{-a.e.} $$
Let $M_y := \{x \in X \mid (x, y) \in M\}$. By Fubini's theorem, $M_y \in \mathcal X$ and $$ \begin{align} (\mu_y \otimes \delta_y) (M) &= \int_X \int_Y 1_{M} (x, z) \mathrm d \delta_y (z) \mathrm d \mu_y (x) \\ &= \int_X 1_{M_y} (x) \mathrm d \mu_y (x) \\ &= \mu_y (M_y). \end{align} $$
It follows that $$ \mu_y (M_y) = 1 \quad \nu\text{-a.e.} $$
Notice that $M_y \subset \pi^{-1} (y)$. This completes the proof.
Update: Below I prove the equality involving $\lambda$. For $C \in \mathcal G$, we define a map $$ f_C: y \to \mathbb R, y \mapsto (\mu_y \otimes \delta_y) (C) . $$
Let $\mathcal C := \{A\times B \mid A \in \mathcal X, B \in \mathcal Y\}$ and $\mathcal D := \{ C \in \mathcal G \mid f_C \text{ is measurable}\}$. If $C = A\times B \in \mathcal C$, then $$ f_C (y) = \left ( \int_X 1_A \mathrm d \mu_y \right ) 1_B(y). $$
By (1.), $y \mapsto \int_X 1_A \mathrm d \mu_y$ is measurable. As such, $f_C$ is measurable for all $C \in \mathcal C$. This implies $\mathcal C \subset \mathcal D$. Clearly, $\mathcal C$ is a $\pi$-system. Let's prove that $\mathcal D$ is a $\lambda$-system. Clearly, $X \times Y \in \mathcal D$. If $C \in \mathcal D$ then $f_{C^c} = 1- f_C$ is measurable and thus $C^c \in \mathcal D$. If $(C_n) \subset \mathcal D$ is pairwise disjoint, then $f_{C} = \sum_n f_{C_n}$ with $C := \bigcup_n C_n$ is measurable and thus $C \in \mathcal D$. By Dynkin's $\pi$−$\lambda$ theorem, we get $\sigma(\mathcal C) \subset \mathcal D$. Hence $f_C$ is measurable for every $C \in \mathcal G$.
Let $\mathcal E := \{C \in \mathcal G \mid \lambda(C) = \int_Y f_C \mathrm d \nu\}$. For $C = A \times B \in \mathcal C$. By (2.), $$ \int_Y f_C \mathrm d \nu = \int_Y \left ( \int_X 1_A \mathrm d \mu_y \right ) 1_B(y) \mathrm d \nu (y) = \int_X 1_A (1_B \circ \pi) \mathrm d \mu = \mu(A \cap \pi^{-1} (B)). $$
On the other hand, $$ \lambda (C) = \mu (\phi^{-1} (C)) = \mu (A \cap \pi^{-1} (B)). $$
As such, $\mathcal C \subset \mathcal E$. Just as above, we can prove that $\mathcal E$ is a $\lambda$-system. By Dynkin's $\pi$−$\lambda$ theorem, we get $\sigma(\mathcal C) \subset \mathcal E$. Hence $\lambda(C) = \int_Y f_C \mathrm d \nu$ for every $C \in \mathcal G$. This completes the proof.