If the formula in the question is exactly how it is written in the book, then this is a bit of slack notation, with the ambiguous looking $f(.)$. The subscripts, while a bit ugly, are one way to make it more clear what the function represents (for $f(.)$ is essentially defined as two different things, which in "nit-picking" world is a contradiction unless $f(.)=0$ or $f(.)=\pm\infty$). In my notation,a capital letter stands for the "random variable" and a small letter for a potential value of that random variable. And the subscripts represent what the distribution is for (e.g. $f_{X|Y}(x|y)$ is the pdf for $X$, given $Y$ evaluated at values $x$ and $y$). I also used a capital $F$ for CDFs and small $f$ for PDFs.
Now I could have just written down "use the jacobian" as an answer, but I can never quite remember exactly what jacobian to use (is it J, or 1 over inverse J, or inverse J?). Going through the tedious process below (of essentially re-deriving the change-of-variables formula for PDFs) is the only way that I feel sure I've got the right answer.
AS far as I can tell, what has occurred here is that you define a "random variable" or "source of uncertainty", and call it $Z_t$. This has an unspecified PDF, denoted by $f_{Z_t}(z_{t})$. I am going to assume that by $D(0,1)$ you mean that the pdf is to have mean 0 and variance 1, but is otherwise ambiguous.
So we write the cumulative distribution function (CDF) for $R_t$ as:
$$F_{R_t|\theta}(r_t|\theta)=Pr(R_t<r_t|\theta)$$
Now $R_t$, given $\theta$ is a 1-to-1 function of $Z_t$. Now we substitute this relation $R_t=\mu_t(\theta)+\sqrt{h_t(\theta)}Z_t$ and re-arrange the expression to get:
$$F_{R_t|\theta}(r_t|\theta)=Pr(\mu_t(\theta)+\sqrt{h_t(\theta)}Z_t<r_t|\theta)=Pr(Z_t<\frac{r_t-\mu_t(\theta)}{\sqrt{h_t(\theta)}}|\theta)$$
Now you can see that this last expression is just the CDF of $Z_t$ evaluated at $\frac{r_t-\mu_t(\theta)}{\sqrt{h_t(\theta)}}$. I assume this is what is meant by $z_t(\theta)$. Now to get from the CDF to the PDF, you need to differentiate with respect to $r_t$.
$$f_{R_t|\theta}(r_t|\theta)=\frac{\partial}{\partial r_t}\left[F_{R_t|\theta}(r_t|\theta)\right]=\frac{\partial}{\partial r_t}\left[F_{Z_t|\theta}(z_t(\theta)|\theta)\right]$$
Now you use the chain rule for differentiation to get:
$$f_{R_t|\theta}(r_t|\theta)=\frac{\partial z_t(\theta)}{\partial r_t}\frac{\partial}{\partial z_t(\theta)}\left[F_{Z_t|\theta}(z_t(\theta)|\theta)\right]=\frac{\partial z_t(\theta)}{\partial r_t}f_{Z_t|\theta}(z_t(\theta)|\theta)$$
Now the derivative here can be calculated as:
$$\frac{\partial z_t(\theta)}{\partial r_t}=\frac{\partial \left[\frac{r_t-\mu_t(\theta)}{\sqrt{h_t(\theta)}}\right]}{\partial r_t}=\frac{1}{\sqrt{h_t(\theta)}}$$
Substituting this back into the answer gives the density required:
$$f_{R_t|\theta}(r_t|\theta)=\frac{f_{Z_t|\theta}(z_t(\theta)|\theta)}{\sqrt{h_t(\theta)}}$$
Note that this actually holds regardless of the distribution of the $Z_t$. For nowhere did I make use of the $D(0,1)$ assumption.
