The same thing is true on Earth. Let's say you plant a flag on the North pole, and then walk some distance $r$ away. The set of all points which are at a distance $r$ from the flag is a circle. What is the circumference of this circle?
The answer isn't $2\pi r$. This can be seen by letting $r = \pi R/2$, which means that you're standing on the equator. The circumference of this circle is then given by $4r$!
More generally, the metric on the 2-sphere is
$$g = \pmatrix{R^2 & 0 \\ 0 & R^2 \sin^2(\theta)}$$
in angular coordinates $(\theta,\phi)$, where $R$ is the radius of the Earth. The infinitesimal distance element is therefore $ds^2 = R^2d\theta^2 + R^2\sin^2(\theta) d\phi^2$. If we walk from the North pole for some distance $r$, we will be moving along a line of constant $\phi$ so the line element is $ds = Rd\theta$, which means that $r=R\theta \iff \theta = r/R$ where $\theta$ is the polar angle corresponding to your chosen circle. The circumference of that circle is obtained by walking along a line of constant $\theta$ from $\phi=0$ to $2\pi$, so $ds = R\sin(\theta) d\phi$ and so the circumference is given by $C = 2\pi R\sin\left(\frac{r}{R}\right)$.
So to summarize, on the surface of the Earth, the circumference of a circle is related to its radius $r$ via
$$C(r) = 2\pi R\sin\left(\frac{r}{R}\right)$$
which is different from the flat space relationship $C_{\mathrm{flat}}(r) = 2\pi r$. If $r\ll R$ then these expressions are approximately the same, which reflects the local apparent flatness of the Earth, but they are ultimately different.
Exactly the same thing is true in the Schwarzschild coordinates around a black hole. The curvature of space is such that the circumference of a circle is no longer simply proportional to the radial distance from its central point. The relationship is more complex than the one presented here, but the concepts are more or less identical - with the exception of some technicalities which would arise if you try to cross the event horizon in these coordinates.
