Why is a gradient field a special case of a vector field?

If you have a gradient field $(X=\partial_xf,Y=\partial_yf,Z=\partial_zf)$,

by using the relation ${\partial^2f\over{\partial x\partial y}}={{\partial^2f}\over{\partial y\partial x}}$ you obtain $\partial_yX=\partial_xY$. This is not true for every vector field.

For example $(x,x^2y^2,z)$ is not a gradient vector field.

A vector field is just a function $\mathbf{F}:\mathbb{R}^n\to\mathbb{R}^n$. (Or the domain can be some subset $D$ of $\mathbb{R}^n$, not all of it, of course. To avoid making this remark repetitively, I'll just stick with $\mathbb{R}^n$, as an example). From an intuitive/geometrical point of view, it is best visualized as having a vector attached to (or starting from) each point of the domain $D$. Given a multivariable function $f:\mathbb{R}^n\to\mathbb{R}$, its gradient $\nabla f$ is precisely that kind of a function, as it acts from $\mathbb{R}^n$ to $\mathbb{R}^n$, so it is a vector field.

What the book says is that not all vector fields can be obtained in this way. Intuitively, it should make sense: the fact that the components of $\nabla f$ are derived from the same original function $f$ should result in some relationships among them. If you put random functions as components, most likely they wouldn't be related in that special way. It's like siblings: being the descendants of the same parents they have some things in common, something that random people wouldn't.

In more mathematical terms, one of such special relationships is the condition of equality of mixed partial derivatives. Say, for a function of two variables $f:\mathbb{R}^2\to\mathbb{R}$, we must have $\displaystyle \frac{\partial^2f}{\partial y\partial x}=\frac{\partial^2f}{\partial x\partial y}$, assuming these derivatives are continuous. Therefore, if $\displaystyle \mathbf{F}=\nabla f=\left\langle\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right\rangle=\langle M,N\rangle$ is its gradient, then we have to have $$\color{blue}{\frac{\partial M}{\partial y}}=\frac{\partial^2f}{\partial y\partial x}=\frac{\partial^2f}{\partial x\partial y}=\color{blue}{\frac{\partial N}{\partial x}}.$$

So being the gradient is very special! This condition allows you to construct vector fields that are not gradients. Pretty much anything you would put into a vector field at random almost surely wouldn't be a gradient. For example, $\mathbf{F}(x,y)=\langle 2x+3y,4x+5y\rangle$ isn't a gradient, because $$\frac{\partial M}{\partial y}=3\neq4=\frac{\partial N}{\partial x}.$$

General vector fields can generate flows and circulations. A gradient fields and only gradient fields (under some additional regularities) always generate circulations that amount to zero. Try to compute the circulation of a gradient field in a plane around an elementary rectangle centered at any point whose sides are parallel to the coordinate axes. You will get an elementary circulation that is zero simply from the fact that mixed second order partial derivatives of the scalar field does not depend on the order of derivation. Notice that the density of such elementary circulations (that is, the curl of the gradient field in every point) is also null.

So gradient fields and only gradient fields (under additional regularities) have curl identically equals to zero.

You can also see that there are fields whose flows (and elementary flow density in every point, that is their divergence) always amount to zero.