$g=\rho^2 (d x^2 + dy^2) \Rightarrow g = ds^2 +\tanh^2 s d\theta^2$ The question is from the 10th page of  Topping's Lectures on Ricci flow, the calculation about  Hamilton's cigar soliton.
Consider $R^2$ with metric
$$
g=\rho^2 (dx^2 + dy^2),~~~\rho^2= \frac{1}{1+x^2 + y^2}
$$
where $dx^2 = dx \otimes dx $. Topping state that
$$
g=ds^2  + \tanh^2 s d\theta^2
$$
in terms of the geodesic distance from the origin $s$, and the polar angle $\theta$. But I can't get it.
What I try: In polar coordinates
$$
x=r\cos\theta,~~~ y = r\sin\theta
$$
I get
$$
g=\rho^2 dr^2 + r^2 \rho^2 d\theta^2
$$
The process:
$$
\begin{pmatrix} 
dx \\
dy 
\end{pmatrix}
=
\begin{pmatrix} 
\cos\theta & -r\sin\theta \\
\sin\theta &  r\cos\theta
\end{pmatrix}
\begin{pmatrix} 
dr \\
d\theta 
\end{pmatrix}
$$
and
\begin{align}
g
&= 
\begin{pmatrix} 
dx &dy 
\end{pmatrix}
\begin{pmatrix} 
\rho^2 & 0 \\
0 & \rho^2 
\end{pmatrix}
\begin{pmatrix} 
dx \\
dy 
\end{pmatrix}
\\
&=
\begin{pmatrix} 
dr &d\theta 
\end{pmatrix}
\begin{pmatrix} 
\cos\theta & \sin\theta \\
-r\sin\theta &  r\cos\theta
\end{pmatrix}
\begin{pmatrix} 
\rho^2 & 0 \\
0 & \rho^2 
\end{pmatrix}
\begin{pmatrix} 
\cos\theta & -r\sin\theta \\
\sin\theta &  r\cos\theta
\end{pmatrix}
\begin{pmatrix} 
dr \\
d\theta 
\end{pmatrix}
\\
&=
\rho^2 dr^2 + r^2 \rho^2 d\theta^2
\end{align}
but I don't know how to deal the geodesic distance $s$.

As you noted, in polar coordinates, the metric reads
$$
g = \frac{1}{1+r^2}\mathrm{d}r^2 + \frac{r^2}{1+r^2}\mathrm{d}\theta^2
$$
Heuristically, you are looking for $s$ such that $\mathrm{d}s^2 = \frac{1}{1+r^2}\mathrm{d}r^2$, which can be stated as ${\mathrm{d}s} = \frac{1}{\sqrt{1+r^2}}{\mathrm{d}r}$. So let's define
$$
s(r) = \int_0^r \frac{1}{\sqrt{1+t^2}}\mathrm{d}t = \sinh^{-1}(r)
$$
It now follows that
$$
\frac{r^2}{1+r^2} = \frac{\sinh(s)^2}{1+\sinh(s)^2} = \frac{\sinh(s)^2}{\cosh(s)^2} = \tanh(s)^2
$$
and finally,
$$
g = \mathrm{d}s^2 + \tanh(s)^2\mathrm{d}\theta^2
$$
