In Schwarzschild coordinates, the change of sign of the $g_{00}$ and $g_{11}$ components of the metric means that, in some sense, $t$ becomes a "spatial" coordinate and $r$ a "temporal" one: the "future" points towards decreasing $r$ instead of increasing $t$, you can see that looking at the light cones in Schwarzschild coordinates, see for example this figure
Schwarzschild light cones in Schwarzschild coordinates (from MTW, page 848)
In addition, inside the surface $r = r_\text{S} = 2M$ you cannot have a positive $\mathrm{d}s^2$ without a non-null $\mathrm{d}r^2$ due to the sign change, so inside a Schwarzschild black hole you have to move. This, again, can be seen using the above light cones: the word line cannot keep constant $r$.
The fact that after having crossed the event horizon light cones point towards the $r = 0$ singularity is true also using other coordinates, such as Kruskal-Szekeres coordinates
Schwarzschild metric in Kruskal-Szekeres coordinates (see the full definition of the coordinates in the Wikipedia article):
\begin{equation}\mathrm{d}s^{2} = \frac{4r_{\text{S}}^{3}}{r}\mathrm{e}^{-r/r_{\text{S}}}(\mathrm{d}v^{2} - \mathrm{d}u^{2}) - r^{2}\mathrm{d}\theta^{2} - r^{2}\sin^{2}\theta\mathrm{d}\phi^{2}\end{equation}
Schwarzschild light cones in Kruskal-Szekeres coordinates. The $r = 0$ region is the one with the inward toothed border (from MTW, page 848)
and Eddington-Finkelstein coordinates
Schwarzschild metric in Eddington-Finkelstein coordinates (see the full definition of the coordinates in the Wikipedia article):
\begin{equation}\mathrm{d}s^{2} = \biggl(1 - \frac{r_{\text{S}}}{r}\biggr)\mathrm{d}\tilde{v}^{2} - 2\mathrm{d}\tilde{v}\mathrm{d}r - r^{2}\mathrm{d}\theta^{2} - r^{2}\sin^{2}\theta\mathrm{d}\phi^{2}\end{equation}
Schwarzschild light cones in Eddington-Finkelstein coordinates (from MTW, page 849)
The sign change has a physical meaning in Schwarzschild coordinates because Schwarzschild $t$ and $r$ coordinates bear physical meanings ($t$ is the far-away time, $r$ the reduced circumference), while I'm not aware of any simple physical meaning of the Kruskal-Szekeres $u$ and $v$ coordinates or of the Eddington-Finkelstein $\tilde{v}$ coordinate. Note that $u$, $v$ and $\tilde{v}$ coordinates mix the original Schwarzschild $t$ and $r$ coordinates. Depending on the coordinates used, there isn't always a sign change in the metric components (in Kruskal-Szekeres coordinates there is no sign change at all), so don't take that change as a general rule.
The Kerr metric tensor, with Boyer-Lindquist coordinates (briefly described in this introduction to the Kerr spacetime by Matt Visser), reads
\begin{equation}
g_{\mu\nu} =
\begin{pmatrix}
(\Delta - a^{2}\sin^{2}\theta)\Sigma^{-1} & 0 & 0 &
a\Sigma^{-1}r_{\text{S}}r\sin^{2}\theta \\
0 & -\Delta^{-1}\Sigma & 0 & 0 \\
0 & 0 & -\Sigma & 0 \\
a\Sigma^{-1}r_{\text{S}}r\sin^{2}\theta & 0 & 0 & -\bigl((r^{2} + a^{2}) +
a^{2}\Sigma^{-1}r_{\text{S}}r\sin^{2}\theta\bigr)\sin^{2}\theta
\end{pmatrix}
\end{equation}
with
\begin{align}
\Delta &= r^{2} - r_{\text{S}}r + a^{2}, \\
\Sigma &= r^{2} + a^{2}\cos^{2}\theta.
\end{align}
The $g_{00}$ component changes its sign on the surfaces
$$r_{\text{E}}^{\pm} = M \pm \sqrt{M^{2} - a^{2}\cos^{2}\theta}$$
Instead, $g_{11}$ changes its sign on the surfaces
$$r_{\pm} = M \pm \sqrt{M^{2} - a^{2}}$$
which determine the outer (with $+$ sign) and the inner (with $-$ sign) event horizons. So they change their sign on two different surfaces. As Jerry Schirmer pointed out in a comment, this would occur also in the Schwarzschild geometry with Kerr-like non-diagonal coordinates (e.g., it occurs with Eddington-Finkelstein coordinates). This doesn't mean that the signature of the metric will change: there always will be a negative (positive) eigenvalue, and three other positive (negative) eigenvalues. In a non-diagonal metric tensor (like Schwarzschild one in Eddington-Finkelstein coordinates or the Kerr one in Boyer-Lindquist coordinates), the diagonal components of the metric tensor aren't necessarily the eigenvalues.
The aim of the isotropic coordinates is to write the metric in the form where the spacelike slices are as close as possible to Euclidean. That is, we try to write the metric in the form:
$$ ds^2 = -A^2(r)dt^2 + B^2(r)d\Sigma^2 $$
where $d\Sigma^2$ is the Euclidean metric:
$$ d\Sigma^2 = dr^2 + r^2(d\theta^2 + \sin^2\theta d\phi^2) $$
So let's use the substitution $r\rightarrow r'$ and write down our metric:
$$ ds^2 = -\left(1-\frac{2M}{r'}\right)dt^2 + B^2(r')\left(dr'^2 + r'^2(d\theta^2 + \sin^2\theta d\phi^2)\right) $$
If we compare this with the Schwarzschild metric:
$$ ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \frac{dr^2}{1 - 2M/r} + r^2(d\theta^2 + \sin^2\theta d\phi^2) $$
Then for the angular parts to be equal we must have:
$$ B^2(r')r'^2 = r^2 $$
And for the radial parts to be equal we must have:
$$ B^2(r')dr'^2 = \frac{dr^2}{1 - 2M/r} $$
Divide the second equation by the first to eliminate $B$ and we end up with:
$$ \frac{dr'^2}{r'^2} = \frac{dr^2}{r^2 - 2Mr} $$
And then just take the square root and integrate and we get the substitution you describe:
$$ r = r'\left(1 + \frac{M}{2r'}\right)^2 $$
Best Answer
If I take the (corrected) formula :
$$R= 2M + \sqrt{r(r-2M)}+2M\ln \left[ \sqrt{\frac{r}{2M} -1} + \sqrt{\frac{r}{2M}} \right] \tag{1}$$
You have (if no error) : $dR = \dfrac{dr}{\sqrt{ 1 - \dfrac{2M}{r}}}$, so $dR^2 = \dfrac{dr^2}{f(r)}$, and this simplifies your metrics.
However, you cannot invert the formula $(1)$ to get $r$ as an explicit function of $R$. You will simply write the metrics :
$$ds^2 = dR^2 + (r(R))^2 \left( d\theta^2 + \sin ^2 \theta \ d\phi\right) \tag{2}$$ where $r(R)$ is implicitely defined by the equation $(1)$