The cylindrical coordinates don't form a vector. They're just a triple of numbers that you can use to describe a point; they don't have properties of vectors such as linearity, e.g. the cylindrical coordinates of the midpoint between two points aren't the average of the cylindrical coordinates of the two points.
What your text is likely referring to by "cylindrical vector" is a representation of a vector in terms of the orthogonal basis given by the unit vectors along $\partial\vec r/\rho$, $\partial\vec r/\phi$ and $\partial\vec r/z$. This is a local basis that depends on $\vec r$, so you're right in saying that your book does a bad job explaining these things; it's ambiguous to ask you to express a vector in this way without specifying the point whose basis is to be used. I'll assume that they mean "express the vector from $C(3,2,-7)$ to $D(-1,-4,2)$ in cylindrical components at $C$". (You omitted the $7$, but it can be reconstructed from the lengths of the vectors.)
Since $\partial\vec r/\partial z$ is just the canonical unit vector in the $z$ direction, the component in that direction is just $2-(-7)=9$. To find the other two components, note that $\partial\vec r/\partial\rho$ points in the direction from the $z$ axis to the point, so the corresponding component is given by
$$
\left(\pmatrix{-1\\-4}-\pmatrix{3\\2}\right)\cdot\pmatrix{3\\2}\Big/\left|\pmatrix{3\\2}\right|=\frac{-24}{\sqrt{13}}\approx-6.66\;.
$$
Then the remaining component along $\partial\vec r/\partial \phi$ is determined up to a sign from the length to be
$$
\pm\sqrt{4^2+6^2-\frac{24^2}{13}}\approx\pm2.77\;,
$$
and the sign is determined by your convention for $\phi$.
$\hat{\theta}$ tells you the angle you need to be at from the positive $x$-axis, and $\hat{r}$ tells you how far you need to walk out from the origin. The magnitude is most certainly given by the $\hat{r}$ component. Perhaps your issue is that you're not adding/multiplying vectors in polar form properly. You cannot simply take, for example, $v_1 = 1\hat{r} + \pi\hat{\theta}$, and conclude that $v_1 + v_1 = 2\hat{r} + 2\pi\hat{\theta}$. Notice that the angle has been changed, which shouldn't happen for two vectors that are colinear.
It is better to represent vectors as $re^{i\theta}$. From which in our example we have, $v_1 + v_1 = e^{i\pi} + e^{i\pi} = 2e^{i\pi}$, which has an $\hat{r}$ of 2, and $\hat{\theta}$ of still $\pi$.
Best Answer
Since $r=\sqrt{x^2+y^2+z^2},$ MathWorld page and your textbook say the same thing. And yes, $z=0$ if and only if $u=v.$
It may help to consider this diagram from the page http://mathworld.wolfram.com/ParabolicCoordinates.html:
This is a cross-section of the system of coordinates showing the curves of constant $u$ and constant $v$ in the $y,z$ plane. One can obtain the complete parabolic coordinate system in $\mathbb R^3$ by rotating the curves in this figure around the $z$ axis, with the result that the points with a given constant $u$ value are all on the surface of a paraboloid, likewise the points with a given constant $v$ value are all on the surface of a paraboloid.
Select an arbitrary positive constant $c$. In the $y,z$ plane, the parabola where $v=c$ is the mirror image (across the $y$ axis) of the parabola where $u=c$. The two parabolas intersect at two points on the $y$ axis. In the full coordinate system, rotating these figures around the $z$ axis, we get a paraboloid where $u=c$ and a mirror-image paraboloid where $v = c$, and the intersection of those paraboloids is a circle in the $x,y$ plane centered at the origin. The radius of that circle is $u^2 = v^2 = c^2$.
If you choose two different constant values of $u$ and $v$, say $u = c_1$ and $v=c_2$ (where $c_1\neq c_2$), the two paraboloids described by $u = c_1$ and $v=c_2$ still intersect in a circle, but the circle is not in the $x,y$ plane. Instead, the circle of intersection is in a plane parallel to the $x,y$ plane at some non-zero value of $z$.
In order to describe a specific point in $\mathbb R^3$ in parabolic coordinates, in addition to the coordinates $u$ and $v$ you need an angle $\theta$, which is an angle of the rotation that turned the parabolas into paraboloids. You can describe a point in the $x,y$ plane in polar coordinates, using $\theta$ as the angle and $u^2$ (or $v^2$, since they are the same in the $x,y$ plane) as the radius.
If you want curvilinear coordinates in two dimensions that are not just a reformulation of polar coordinates, you might want to try parabolic cylindrical coordinates instead (http://mathworld.wolfram.com/ParabolicCylindricalCoordinates.html) and simply delete the $z$ axis. The resulting coordinate system looks like the figure above, except that the vertical axis is labeled $x$ instead of $z$. If you would prefer to have your $x,y$ coordinates in a more conventional orientation (I would), simply reflect the figure through the line $x=y$ (that is, flip it over so the axes are where you want them). Note that there is an ambiguity in the coordinate system because the parabolas for $u=u_1$ and $v=v_1$ (for constants $u_1$ and $v_1$) intersect in two places, one with a positive $y$ coordinate and one with a negative $y$ coordinate. As long as you are looking within a region that is all on one side of the $x$ axis, this is not a problem.