Why do disks on planes grow more quickly with radius than disks on spheres? In the book, Mr. Tompkins in Wonderland, there is written something like this:

On a sphere the area within a given radius grows more slowly with the radius than on a plane.

Could you explain this to me? I think that formulas shows something totally different:
The area of a sphere is $4 \pi r^2$, but the area of a disk on the plane is $\pi r^2$.
UPDATE:
Later in this book there is:

If, for example, you are on the north pole, the circle with the radius equal to a half meridian is the equator, and the area included is the northern hemisphere. Increase the radius twice and you will get in all the earth's surface; the area will increase only twice instead of four times if it were on a plane.

Could someone explain me this?

Theorem: Let $0 < r \leq 2R$ be real numbers, $S_{1}$ a Euclidean sphere of radius $R$, $S_{2}$ a Euclidean sphere of radius $r$ centered at a point $O$ of $S_{1}$, and $D_{r}$ the portion of $S_{1}$ inside $S_{2}$.
The area of $D_{r}$ is $\pi r^{2}$.

Now let $D'_{r}$ be the disk of radius $r$ in $S_{1}$, i.e., the portion of $S_{1}$ at distance at most $r$ from $O$, measured along the surface of $S_{1}$. Since a chord of length $r$ subtends an arc of length strictly greater than $r$, the disk $D_{r}'$ is strictly smaller than $D_{r}$, and therefore has smaller area.

Proof of Theorem: The region $D_{r}$ (bold arc) is a zone on a sphere of radius $R$. To calculate its width, consider a longitudinal section:

Let $\theta$ be the angle subtended at the center of $S_{1}$ by a chord of length $r$. The indicated right triangles share an interior angle, and consequently are similar, with $\frac{\theta}{2}$ as interior angle. Reading from the right triangle whose hypotenuse is horizontal,
$$
R\sin\tfrac{\theta}{2} = \tfrac{r}{2}.
$$
On the other hand, reading from the right triangle with the dashed vertical side shows that $D_{r}$ has width $h = r\sin\frac{\theta}{2}$. By Archimedes' theorem, the area of $D_{r}$ is
$$
2\pi Rh = 2\pi R \cdot r\sin\tfrac{\theta}{2}
  = 2\pi r \cdot R\sin\tfrac{\theta}{2}
  = \pi r^{2}.
$$
