Which of the two $k$ values do I use in solving for $r'$ and $z'$ ?
Either one will be a point of intersection, since in general a line intersects an ellipse in two points. Which one you need, or if it doesn't matter, depends on your application.
How can you tell when the line doesn't intersect the ellipse? Is the quadratic equation for $k$ not have any real roots?
Right. And if you only have a single root, i.e. both roots coincide, then the line will be a tangent.
Is it important for $r$ to be $>0$ in for this equation even though I'm not testing a point, but a line instead?
I see no reason to require this here.
Why is the semi-minor axis being defined as $a_e(1-f)$? I usually just define the semi-minor axis the same way I define the major, so could I just replace all the $(1-f)$s in the above equations with my desired semi-minor axis length?
Yes, you can write the semi-minor axis (called $a_p$ in your document) instead of $a_e(1-f)$. Which means you could replace the $(1-f)$ themselves by $\frac{a_p}{a_e}$ or multiply the whole equation by $a_e^2$.
Finally, is there any simpler, faster way to see if and were an ellipse and line intersect?
Depends on how your objects are given. If you really have an ellipse and a line in the described form, I can't think of something fundamentally easier. If, on the other hand, you have an ellipse in some other representation, then you might be able to use that representation directly to compute the intersection, instead of transforming its representation first. In any case, you can't avoid the quadratic equation.
The distance should be easy to determine in closed-form from $r$ (cylinder radius), $\overline{C}$ (a vector along the cylinder's center) and $\overline{P}$ (a vector representing a ray from that centerline to the point $p$).
My first stab...
$$ d = r \tan\left(\arccos{\left(\! \frac{\overline{C}\cdot\overline{P}}{|\overline{C}| |\overline{P}|} \!\right)}\right) $$
Best Answer
If your cylinder is set along an an axis, one way you could think of this is the following:
You "watch" your cylinder with your vision axis colinear with the cylinder axis, you can transform your parameterized line in the same referential as the cylinder then solve for a simple circle-line intersection in 2D.
I'd recommend working with a paremeterized line too, then finding:
You are not forced to solve quadratics in order to find this intersection, but you can't avoid some trigonometry, here is an example: