Tortoise coordinate transformation The differential form $dr$ can be written $\left(1-\dfrac{2GM}{r}\right)dr^*$ where $r^*$ is the tortoise coordinate. Writing the Schwartzchild metric then gives
$ds^2$ = $\left(1-\dfrac{2GM}{r}\right)\left(dt^2 - dr^{*2}\right)$
The problem is that we are dealing with $dr^2$ in the schwartzchild metric, so shouldn’t we do $drdr$ giving us $dr^2$ = $\left(1-\dfrac{2GM}{r}\right)^2dr^{*2}$
So why don’t we have $\left(1-\dfrac{2GM}{r}\right)^2$ in the metric in terms of the tortoise coordinate?
 A: $ds^2 = f(r)~dt^2 - \dfrac{dr^2}{f(r)} - r^2~d\Omega^2$ ; $f(r) = 1-\dfrac{2GM}{r}$
To avoid coordinate singularity at $r=2GM$, we don't want $f(r)$ in the denominator in the second term. So take common $f(r)$ from first two terms:
$$ds^2 = f(r)~\left(dt^2 - \dfrac{dr^2}{f(r)^2}\right) - r^2~d\Omega^2$$
Declare $\dfrac{dr}{f(r)}$ as $dr^*$, so that $\dfrac{dr^2}{f(r)^2}$ becomes simply $dr^{*2}$ and horizon lies at $r^* \longrightarrow -\infty$ and infinity simply at $r^* \longrightarrow +\infty$ (we want to describe the region outside the horizon of the spherical body i.e. $r \gt 2GM$); Hence,
$$ds^2 = f(r)~\left(dt^2 - dr^{*2}\right) - r^2~d\Omega^2$$
where $r^* = r+2M ~\ln~\left|\dfrac{r}{2GM}-1\right|$ in natural choice of units, $c=1$.
[If you want to know that why we write $f(r) = \left(1-\dfrac{2GM}{r}\right)$ in terms of $r$, not in terms of $r^*$, while we are giving an coordinate transformation $r \longrightarrow r^*$ (I'm not quite sure about the intended sense of your question), then go through the following.]
Therefore,
$$e^{r^*} = e^r ~\left|\dfrac{r}{2GM}-1\right|^{2GM}$$
$$\Longrightarrow e^{\frac{r^*}{2GM}} = e^{\frac{r}{2GM}} ~\left(\dfrac{r}{2GM}-1\right)$$
we are interested in the region $r \gt 2GM$ (and hence the tortoise coordinate). So,
$$e^{\frac{r^*}{2GM}-1} = e^{\frac{r}{2GM}-1} ~\left(\dfrac{r}{2GM}-1\right)$$
$$\Longrightarrow \dfrac{r}{2GM} = 1+W\left(e^{\frac{r^*}{2GM} - 1}\right)$$
$$\Longrightarrow r = 2GM \left(1+W\left(e^{\frac{r^*}{2GM} - 1}\right)\right)$$ where $W$ is the Lambert W function.
So your $f(r)$ becomes, $\left(1-\dfrac{1}{1+W\left(e^{\frac{r^*}{2GM} - 1}\right)}\right)$; an unnecessarily complicated expression! Do you want to write such a cumbersome expression all the time? It is often hard to handle and manipulate.
But if you express $f(r)$ in term of ordinary Schwarzschild coordinates $r$, without loss of any generality, your expressions and equations will be easier to tackle. So...
Moreover probably you know, the coordinate transformation $r \longrightarrow r^*$ is done not to compactify the metric in a shorter way, but

*

*to remove the coordinate singularity at the horizon, $r \longrightarrow 2GM ~\left(r^* \longrightarrow -\infty \right)$ and

*to express only the exterior of the black hole (or that spherically symmetric object) in a more convenient and useful way.

So you only need to replace $\dfrac{dr}{f(r)}$ by $dr^*$ to fulfill the aforementioned needs. Whether or not you are modifying your $f(r)$ by $r^*$, is not at all important in most of the physical cases!
