Let $X$ be an absolutely continuous real-valued random variable whose distribution is $\mu_X$ and whose p.d.f. is $p_X$. Let $f:\mathbb R \to \mathbb R$ be differentiable such that $f'(t)>0$ for all $t\in \mathbb R$. This implies $f$ is strictly increasing. Let $Y := f(X)$. Clearly, the distribution $\mu_Y$ of $Y$ is absolutely continuous w.r.t. Lebesgue measure $\lambda$. Let $F_Y,p_Y$ be the c.d.f. and p.d.f. of $Y$ respectively.
Clearly, $\mu_Y$ is absolutely continuous w.r.t. $\mu_X$. According to this Wikipedia page, $$ p_Y := \frac{\mathrm d \mu_Y}{\mathrm d \lambda} = \frac{\mathrm d \mu_Y}{\mathrm d \mu_X} \frac{\mathrm d \mu_X}{\mathrm d \lambda}. $$
Here $\frac{\mathrm d \mu_Y}{\mathrm d \mu_X}$ is the Radon–Nikodym derivative of $\mu_Y$ w.r.t. $\mu_X$. Similarly, $\frac{\mathrm d \mu_X}{\mathrm d \lambda}$ is the Radon–Nikodym derivative of $\mu_X$ w.r.t. $\lambda$. At page $14$ of this lecture note, the author mentioned that
$$ p_Y (t) = \frac{\mathrm d F_Y (t)}{\mathrm d t} = \frac{\mathrm d F_X (f^{-1} (t))}{\mathrm d t} \frac{\mathrm d f^{-1} (t)}{\mathrm d t} =\frac{p_X (f^{-1} (t))}{f'(f^{-1} (t))}. $$
Clearly, the map $F_\lambda:\mathbb R \to \mathbb R, t \mapsto t$ is the corresponding c.d.f. of $\lambda$. Obviously, the p.d.f. of $\lambda$ is $p_\lambda:\mathbb R \to \mathbb R, t \mapsto 1$. It seems to me the author meant by $\frac{\mathrm d F_X (f^{-1} (t))}{\mathrm d t}$ the value of $p_X :=\frac{\mathrm d \mu_X}{\mathrm d \lambda}$ at $f^{-1} (t)$, i.e., $$ \frac{\mathrm d F_X (f^{-1} (t))}{\mathrm d t} = \frac{\mathrm d \mu_X}{\mathrm d \lambda} (f^{-1} (t)) =p_X (f^{-1} (t)). $$
Could you elaborate on how to transfer from the Radon–Nikodym derivative $\frac{\mathrm d \mu_Y}{\mathrm d \mu_X}$ to the classical derivative $\frac{\mathrm d f^{-1} (t)}{\mathrm d t}$, i.e., $$ \frac{\mathrm d \mu_Y}{\mathrm d \mu_X} (t) = \frac{\mathrm d f^{-1}}{\mathrm d t} (t)? $$
