$\newcommand{\R}{\mathbb R}
\newcommand{\B}{\mathcal B}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\renewcommand{\c}{\circ}
\newcommand{\tr}{\operatorname{tr}}$
The desired function $f$ and random variable (r.v.) $G$ can be built recursively, by induction, using the increasing rearrangement/inverse transformation method: If $F$ is any cumulative distribution function (cdf),
\begin{equation}
F^{-1}(u):=\inf\{x\in\mathbb R\colon F(x)\ge u\}
\end{equation}
for $u\in(0,1)$, and $U\sim\mathcal U(0,1)$ (a r.v. uniformly distributed on the interval $(0,1)$), then the cdf of the r.v. $F^{-1}(U)$ is $F$.
Indeed, for $j=0,\dots,n$, let $\pi_j$ be the push-forward image of the probability measure $\pi$ under the projection of $\R^n\times\R^n$ onto $\R^n\times\R^j$, so that $\pi_j(A\times B_j)=\pi(A\times B_j\times\R^{n-j})$ for $A$ in the Borel sigma-algebra $\B(\R^n)$ and $B_j$ in $\B(\R^j)$; naturally, $\R^0=\{0\}$, and we identify $\R^j\times\R^{n-j}$ with $\R^n$.
Similarly, let $\nu_j$ be the push-forward image of the probability measure $\nu$ under the projection of $\R^n$ onto $\R^j$, so that $\nu_j(B_j)=\nu( B_j\times\R^{n-j})$ for $B_j$ in $\B(\R^j)$.
Then $\pi_0 =\mu$, $\pi_n=\pi$, $\nu_0$ is the only probability measure on $\B(\R^0)=\B(\{0\})$, and $\nu_n=\nu$.
Write $Y=(Y_1,\dots,Y_n)$ and let $Y_{1;j}:=(Y_1,\dots,Y_j)$, with $Y_{1;0}:=0$; the r.v.'s $Y_1,\dots,Y_n$ are to be constructed.
Accordingly, for $y=(y_1,\dots,y_n)\in\R^n$ let $y_{1;j}:=(y_1,\dots,y_j)$, with $y_{1;0}:=0$. For $j=0,\dots,n$, let $G_j:=(U_1,\dots,U_j)$, where $U_1,\dots,U_n$ are independent $\mathcal U(0,1)$ r.v.'s. In particular, $G_0=0$.
For $j=1,\dots,n$, we are going to construct, by induction, a function $f_j\colon\R^n\times(0,1)^j\to\R^j$ such that for $Y_{1;j}:=f_j(X,G_j)$ the distribution of $(X,Y_{1;j})$ is $\pi_j$ and hence the distribution of $Y_{1;j}$ is $\nu_j$.
To complete the basis of induction, let $f_0(x,0):=0$ for all $x\in\R^n$, so that $Y_0=f_0(X,G_0)$.
Take now any $j=1,\dots,n$. Let
$$\R^n\times\R^{j-1}\times\B(\R)\ni(x,y_{1;j-1},C)\longmapsto
\la_{x,y_{1;j-1}}(C)$$
be a regular version of the conditional distribution of $Y_j$ given $(X,Y_{1;j-1})$ assuming that the joint distribution of $(X,Y_{1;j})$ is $\pi_j$, so that
$$\pi_j(dx\times dy_{1;j-1}\times dy_j)=\pi_{j-1}(dx\times dy_{1;j-1})\la_{x,y_{1;j-1}}(dy_j).$$
For each $(x,y_{1;j-1})\in\R^n\times\R^{j-1}$, let $F_{x,y_{1;j-1}}$ be the cdf of the probability measure $\la_{x,y_{1;j-1}}$ on $\B(\R)$, and define the function $f_j\colon\R^n\times(0,1)^j\to\R^j$ by the formula
\begin{equation}
f_j(x,u_{1;j}):=\big(y_{1;j-1},F^{-1}_{x,y_{1;j-1}}(u_j)\big)\quad\text{with}\quad
y_{1;j-1}=f_{j-1}(x,u_{1;j-1})
\end{equation}
for $(x,u_{1;j})\in\R^n\times(0,1)^j$, where the notation $u_{1;j}$ is of course quite similar to $y_{1;j}$.
Let now $Y_{1;j}:=f_j(X,U_{1;j})=f_j(X,G_j)$, which is in agreement with the definition $Y_{1;j-1}:=f_{j-1}(X,U_{1;j-1})=f_{j-1}(X,G_{j-1})$ at the previous step of the induction process.
Then the distribution of $(X,Y_{1;j})$ is $\pi_j$ and hence the distribution of $Y_{1;j}$ is $\nu_j$.
In particular, the distribution of $(X,Y)=(X,Y_{1;n})$ is $\pi_n=\pi$ and hence the distribution of $Y=Y_{1;n}$ is $\nu$. Moreover, $Y=Y_{1;n}=f_n(X,G_n)$, as desired.
