$(ct, r_0, \phi_0, z) \text{ for an observer on the rotating disk}$

I think you misread or mistranslated this part. These are the coordinates of non rotating inertial observer at the centre of the disk.

$ds^2 = -c^2dt^2 + dr_0^2 + r_0^2d\phi_0^2$

We can replace $ds^2$ with $-c^2 \tau^2$ where $\tau$ is the proper time of the observer that is rotating with the disk. See Wikipedia so:

. $-c^2 \tau^2 = -c^2dt^2 + dr_0^2 + r_0^2d\phi_0^2$

$\frac{d\tau^2}{dt^2} = 1 - \frac{dr_0^2}{c^2 dt^2} - \frac{r_0^2 d\phi_0^2}{c^2 dt^2}$

For motion at a constant radius $dr_0 =0$ and since angular velocity $\omega = d\phi/ dt$ and tangential velocity $v = r_0 \omega$ we can write:

$\frac{d\tau^2}{dt^2} = 1 - \frac{r_0^2 \omega^2}{c^2} = 1 - \frac{v^2}{c^2}$

$\frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}} $

Looks familiar? The clock rotating at radius $r_0$ is ticking slower than the non rotating clock of the inertial observer. I.e. the rotating clock and the lab frame clock (if the lab frame is the inertial frame in this case - you did not make it clear) are not ticking at the same rate.

Using the above result you get that the circumference C is $C=\gamma(ωr) 2 \pi r$. How is this a contraction if this is greater than 2πr ?

It not the circumference of the disk as whole that is contracting. It is the rulers of the observer riding with the rim. Because her rulers are contracted, she can fit more of them around the perimeter and and the circumference appears larger to her.

$(ct, r_0, \phi_0, z) \text{ for an observer on the rotating disk}$

I think you misread or mistranslated this part, or the book is in error. These are the coordinates of non rotating inertial observer at the centre of the disk.

$ds^2 = -c^2dt^2 + dr_0^2 + r_0^2d\phi_0^2$

We can replace $ds^2$ with $-c^2 \tau^2$ where $\tau$ is the proper time of the observer that is rotating with the disk. See Wikipedia so:

. $-c^2 \tau^2 = -c^2dt^2 + dr_0^2 + r_0^2d\phi_0^2$

$\frac{d\tau^2}{dt^2} = 1 - \frac{dr_0^2}{c^2 dt^2} - \frac{r_0^2 d\phi_0^2}{c^2 dt^2}$

For motion at a constant radius $dr_0 =0$ and since angular velocity $\omega = d\phi/ dt$ and tangential velocity $v = r_0 \omega$ we can write:

$\frac{d\tau^2}{dt^2} = 1 - \frac{r_0^2 \omega^2}{c^2} = 1 - \frac{v^2}{c^2}$

$\frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}} $

Looks familiar? This is the concept of "Instantaneous Comoving Observer" at work. The time dilation of a clock following a circular path and experiencing centripetal acceleration is exactly the same as an inertial clock moving in a straight line, when measured in a small enough spatial and time interval. The clock rotating at radius $r_0$ is ticking slower than the non rotating clock of the inertial observer. I.e. the rotating clock and the lab frame clock (if the lab frame is the inertial frame in this case - you did not make it clear) are not ticking at the same rate.

Using the above result you get that the circumference C is $C=\gamma(ωr) 2 \pi r$. How is this a contraction if this is greater than 2πr ?

It not the circumference of the disk as whole that is contracting. It is the rulers of the observer riding with the rim. Because her rulers are contracted, she can fit more of them around the perimeter and and the circumference appears larger to her.