Radon-Nikodym derivative of pushforward of Lebesgue measure by differentiable function with respect to Lebesgue measure

Your intuition makes no sense - the Jacobian is at a point of the domain, and the Radon-Nikodym derivative is at a point in the target.

Notice this Radon-Nikodym derivative exists [...] in this setup.

Not really. As was pointed out, the constant map pushes forward to an "infinite-sized delta" at the value. This has no Radon-Nikodym derivative with respect to Lebesgue, both because the pushforward is not sigma-finite (hence "infinite-sized") and because it is not absolutely continuous relative to Lebesgue (hence "delta").

A simple case with a decent answer is when $f$ is a proper submersion. Then at $y\in \mathbb{R}^m$ we have $M=f^{-1}(y)$ a smooth compact submanifold of $\mathbb{R}^n$, and the derivative $\frac{d f_*\lambda^n}{d\lambda^m}(y)$ is the integral over $M$ of the multidimensional Jacobian you mention (note that this reduces to the correct answer when $f$ is a diffeo). The reason is that the multidimensional Jacobian is the infinitesimal volume stretching factor between tangent spaces on domain and target. In other words, a preimage of a small neighborhood of $y$ of volume $V$ is a thickening of $M$, fibered over $M$ by orthocomplements to the kernel of $f$ at every $x\in M$. The volume of each such orthocomplement is approximately $V\times Jf(x)$, so that the overall volume of the thickening is approximately $V\times\int_M (Jf(x)) d(vol M)$. The Radon-Nikodym derivative is the limit of the ratio of that to $V$ when $V$ becomes small.

If $U, V \subset \mathbb{R}^n$ are open sets and $f: U \to V$ is a smooth injective map with non-vanishing Jacobian determinant $J_f$, then the pushforward $\lambda \circ f^{-1}$ (denoted $f_*\lambda$ in the original question) of the Lebesgue measure $\lambda$ is absolutely continuous with respect to $\lambda$ (restricted to the $U$ and $V$, respectively) and it is given by:

$$ (\lambda \circ f^{-1})(\mathrm{d}y) = \left[J_f(f^{-1}(y)) \right]^{-1} \lambda(\mathrm{d}y) = J_{f^{-1}}(y) \lambda(\mathrm{d}y). \tag{1} $$

In other words, the term $J_{f^{-1}}(y)$ is the density of the pushforward measure $f_*\lambda$ with respect to the Lebesgue measure $\lambda$:

$$ \frac{(\lambda \circ f^{-1})(\mathrm{d}y)}{\lambda(\mathrm{d}y)} = J_{f^{-1}}(y). \tag{2} $$

More generally, if we have any measure $\mu(\mathrm{d}x) = h(x) \lambda(\mathrm{d}x)$ that is absolutely continuous with respect to $\lambda$, then

$$ \frac{(\mu \circ f^{-1})(\mathrm{d}y)}{\lambda(\mathrm{d}y)} = h(f^{-1}(y)) J_{f^{-1}}(y). \tag{3} $$

This equation corresponds to the usual rule that a derived random variable $Y = f(X)$ for 1-1 function $f$ has density

$$ p_Y(y) = p_X(f^{-1}(y))J_{f^{-1}}(y). $$

Theorem 17.2 of Billingsley states a related property; namely, if $f: U \to V$ is defined as above and $\phi: V \to R$ is a Borel function, then

$$ \int_{V} \phi(x) \lambda(\mathrm{d}x) = \int_{U} \phi(f(x)) \underbrace{J_f(x)\lambda(\mathrm{d}x)}_{:=\ \nu(\mathrm{d}x)} \tag{4} $$

All that is left now is to do pattern matching on the definition of the pushforward measure. Namely, with $\nu$ as defined above, consider the pushforward measure $(\nu \circ f^{-1})$; we have:

$$ \int_{V}\phi(x) (\nu \circ f^{-1})(\mathrm{d}x) = \int_{U}\phi(f(x))\nu(\mathrm{d}x) \tag{*} $$

Therefore, the pushforward of the measure $\nu(dx) = J_f(x) \lambda(\mathrm{d}x)$ through $f$ (denoted $\nu \circ f^{-1})$ is exactly the Lebesgue measure $\lambda$. This result is unsurprising, since applying $(3)$ we have

$$ \begin{align} \frac{(\nu \circ f^{-1})(\mathrm{d}y)}{\lambda(\mathrm{d}y)} &= J_f(f^{-1}(y)) J_{f^{-1}}(y)\\ &= J_f(f^{-1}(y)) \left[J_f(f^{-1}(y)) \right]^{-1} \\ &= 1. \end{align} $$

It should be noted that $(*)$ is a very special property of the Lebesgue measure, and it is the essential equation that is needed to obtain equations $(1)$ through $(3)$, and not the other way round, which is why Billingsley states it as the key theorem.