"but how can you have a "basis" that changes at every point?"
This is really the root of your problem. Mathematically (and physically) speaking, such a basis works fine. You just have a (hopefully temporary) conceptual problem.
Maybe try thinking of it this way: How does $\hat{x}$ know to point "to the right" in cartesian coordinates, at an arbitrary point in the coordinate system? You could define it as "parallel to the $x$-axis". Similarly for $\hat{y}$. Now how would you define $\hat{r}$? Parallel to a line joining the current point and the origin. $\hat{\theta}$? Perpendicular to $\hat{r}$. Not that different, right?
You were pretty close already. There is a handy table on Wikipedia for a variety of coordinate systems. But for the polar system:
$$ \vec{\nabla} \cdot \vec{U} = \frac{\partial U_r}{\partial r} + \frac{1}{r} \frac{\partial U_\theta}{\partial \theta} $$
and you can look up the curl in the same table.
These can be derived from the Cartesian definitions by considering the total differentials:
$$ dr = \frac{\partial r}{\partial x} dx + \frac{\partial r}{\partial y} dy$$
and
$$ d\theta = \frac{\partial \theta}{\partial x} dx + \frac{\partial \theta}{\partial y} dy$$
Or to put it another way, you end up with:
$$ \frac{\partial \phi}{\partial r} = \frac{\partial \phi}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial \phi}{\partial y}\frac{\partial y}{\partial r}$$
which is just the chain-rule for derivatives. $\phi$ is any variable. A similar expression can be written for the $\theta$ coordinate. You can use this to go between any coordinate systems you would like to go between and relate it all back to the Cartesian system by successive chaining.
As for a physical insight, that's a little bit harder. But the way I like to look at it for the polar (aka cylindrical) system is that the unit length of $\theta$ changes as $r$ changes. If you draw a big circle vs a small circle, the arc length of the wedge defined by an angle $\theta$ is certainly different. And so it is natural that the derivative in the $\theta$ direction will have some radial dependence.
Best Answer
As the text says, $\frac{\partial {\bf{r}}}{\partial r}$ is a vector which points in the $\hat r$ direction. If this is not clear, consider the difference quotient:
$$\frac{\partial {\bf{r}}}{\partial r} \equiv \lim_{\delta r \rightarrow 0} \frac{{\bf{r}}(r+\delta r,\theta,\phi) - {\bf{r}}(r,\theta,\phi)}{\delta r}$$
The numerator is an arrow which points from a point $(r,\theta,\phi)$ to the point $(r+\delta r,\theta,\phi)$, and so it points in the $\hat r$ direction. There is no particular reason that this should be a unit vector, though, so if we want the radial unit vector, we should say that
$$\hat r = \frac{\frac{\partial {\bf{r}}}{\partial r}}{\left|\frac{\partial {\bf{r}}}{\partial r}\right|}$$
Precisely the same reasoning holds for $\hat \theta$ and $\hat \phi$.