Goal. I want to know if the proposition below is true when $\beta(x,y)$ is not differentiable in $x$. If the proposition is false, I'd like a counterexample.
Setup. Set $X=\mathbb{R}^m$ and $Y=\mathbb{R}^n$. Assume $\alpha:X\to X$ is a $C^1$-diffeomorphism and $\beta:X\times Y\to Y$ satisfies the following: For each $x\in X$, the map $\beta_x:Y\to Y$ given by $\beta_x(y)=\beta(x,y)$ is a $C^1$-diffeomorphism. Define $f:X\times Y\to X\times Y$ by $f(x,y)=(\alpha(x),\beta_x(y))$. Note $f$ has inverse $f^{-1}(x,y)=(\alpha^{-1}(x),\beta_{\alpha^{-1}(x)}^{-1}(y))$.
Proposition. Let $\pi(x,y)=\pi_2(y\mid x)\pi_1(x)$ be a probability density function (assume continuous/$C^1$/smooth as needed) on $X\times Y$. Then \begin{align*} f_*\pi(x,y) &= (\beta_{\alpha^{-1}(x)})_*\pi_2(y\mid\alpha^{-1}(x))\alpha_*\pi_1(x) \end{align*} where $f_*\pi$ and $\alpha_*\pi_1$ are the pushforward densities of $\pi$ and $\pi_1$ by $f$ and $\alpha$, respectively, and $(\beta_x)_*\pi_2(y\mid x)$ is the pushforward of the conditional probability density $\pi_2(\cdot\mid x)$ on $Y$ by $\beta_x$.
Attempted proof. The differential of $f^{-1}(x,y)=(\alpha^{-1}(x),\beta_{\alpha^{-1}(x)}^{-1}(y))$ is \begin{align*} df^{-1}(x,y) = \begin{pmatrix} \partial_x\alpha^{-1}(x) & 0 \\ \partial_x\beta^{-1}_{\alpha^{-1}(x)}(y) & \partial_y\beta^{-1}_{\alpha^{-1}(x)}(y) \end{pmatrix}. \end{align*} Hence $\lvert\det df^{-1}(x,y)\rvert=\lvert\det \partial_x\alpha^{-1}(x)\rvert\lvert\det\partial_y\beta^{-1}_{\alpha^{-1}(x)}(y)\rvert$. So by change of variables, \begin{align*} f_*\pi(x,y) &= \pi(\alpha^{-1}(x),\beta^{-1}_{\alpha^{-1}(x)}(y))\lvert\det \partial_x\alpha^{-1}(x)\rvert\lvert\det\partial_y\beta^{-1}_{\alpha^{-1}(x)}(y)\rvert \\ &= \pi_2(\beta^{-1}_{\alpha^{-1}(x)}(y)\mid\alpha^{-1}(x))\pi_1(\alpha^{-1}(x))\lvert\det \partial_x\alpha^{-1}(x)\rvert\lvert\det\partial_y\beta^{-1}_{\alpha^{-1}(x)}(y)\rvert \\ &= (\beta_{\alpha^{-1}(x)})_*\pi_2(y\mid\alpha^{-1}(x))\alpha_*\pi_1(x). \end{align*}
Problem. In using change of variables I've assumed $\beta^{-1}_{\alpha^{-1}(x)}(y)$ can be differentiated in $x$; I don't want to make this assumption in general. Can this be relaxed, especially given that $\partial_x\beta^{-1}_{\alpha^{-1}(x)}(y)$ doesn't contribute to the determinant of the above matrix?
The proposition is true. Just use Fubini's theorem. Specifically, let $\Pi$ denote the density with density $\pi$. Then for any bounded, measurable function $\phi:X\times Y\to\mathbb{R}$, \begin{align*} f_*\Pi(\phi) &= \int_Y\int_X (\phi\circ f)(x,y)\pi_2(y\mid x)\pi_1(x)dxdy \\ &= \int_Y\int_X \phi(\alpha(x),\beta_x(y))\pi_2(y\mid x)\pi_1(x)dxdy \\ &= \int_Y\int_X \phi(u,\beta_{\alpha^{-1}(u)}(y))\pi_2(y\mid\alpha^{-1}(u))\pi_1(\alpha^{-1}(u))\lvert\det d\alpha^{-1}(u)\rvert dudy \\ &= \int_X\int_Y \phi(u,\beta_{\alpha^{-1}(u)}(y))\pi_2(y\mid\alpha^{-1}(u))dy\ \alpha_*\pi_1(u)du \\ &= \int_X\int_Y \phi(u,v)\pi_2(\beta^{-1}_{\alpha^{-1}(u)}(v)\mid\alpha^{-1}(u))\lvert\det d_v\beta^{-1}_{\alpha^{-1}(u)}(v)\rvert dv\ \alpha_*\pi_1(u)du \\ &= \int_X\int_Y\phi(u,v)(\beta_{\alpha^{-1}(u)})_*\pi_2(v\mid\alpha^{-1}(u))\alpha_*\pi_1(u)dvdu. \end{align*} In the third equality I made the substitution $u=\alpha(x)$. In the fifth I made the substitution $v=\beta_{\alpha^{-1}(u)}(y)$.