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.
Here's a reasonable method: translate everything such that the center of the ellipse is at the origin. Consider the intersection of the ellipse with major axis $2a$ and minor axis $2b$ with the polar equation
$$r=\frac{a b}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}$$
and the line $\tan\,\theta=\dfrac{y_2-y_1}{x_2-x_1}$. (When solving the last equation for $\theta$, you will want to use the two-argument arctangent that is implemented in most computing environments.) Once having computed the corresponding values of $r$ at $\theta$ and $\pi+\theta$, convert to rectangular coordinates and translate back to your initialal origin.
Best Answer
First, I assume the ellipse you're using has a center at $(x_0,y_0)$, a width of $2a$ and a height of $2b$. If so, the entire equation then is
$$\frac{(x-x_{0})^2}{a^2} + \frac{(y-y_{0})^2}{b^2} = 1 \tag{1}\label{eq1A}$$
You also give
$$y = \tan(\theta) \cdot x + n \tag{2}\label{eq2A}$$ $$n = y_{0} - x_{0} \cdot \tan(\theta) \tag{3}\label{eq3A}$$
Substitute \eqref{eq3A} into \eqref{eq2A}, rearrange and factor to get
$$y - y_0 = \left(x - x_0\right)\tan(\theta) \tag{4}\label{eq4A}$$
This equation indicates $\theta$ is the angle, relative to the positive $x$-axis, of the line through the ellipse center.
At the intersection points (note there will be $2$ of them), the values of $x$ and $y$ will need to satisfy both \eqref{eq1A} and \eqref{eq4A} simultaneously. To determine their values, substitute \eqref{eq4A} into \eqref{eq1A} to get
\begin{align} 1 & = \frac{(x-x_{0})^2}{a^2} + \frac{\left((x-x_{0})\tan(\theta)\right)^2}{b^2} \\ & = (x-x_{0})^2\left(\frac{1}{a^2} + \frac{\tan^2(\theta)}{b^2}\right) \\ & = (x-x_{0})^2\left(\frac{b^2 + a^2\tan^2(\theta)}{a^2 b^2}\right) \tag{5}\label{eq5A} \end{align}
To make the handling a bit simpler, let
$$k = \frac{a^2 b^2}{b^2 + a^2\tan^2(\theta)} \tag{6}\label{eq6A}$$
since $k$ is a constant value. Next, multiplying both sides in \eqref{eq5A} by $k$ and then taking square roots gives
$$x - x_0 = \pm\sqrt{k} \;\;\to\;\; x = x_0 \pm\sqrt{k} \tag{7}\label{eq7A}$$
Substituting \eqref{eq7A} into \eqref{eq4A} then results in
$$y - y_0 = \pm \sqrt{k} \tan(\theta) \;\;\to\;\; y = y_0 \pm \sqrt{k} \tan(\theta) \tag{8}\label{eq8A}$$
Thus, the $2$ points of intersection between the line and the ellipse are $\left(x_0 + \sqrt{k}, y_0 + \sqrt{k} \tan(\theta)\right)$ and $\left(x_0 - \sqrt{k}, y_0 - \sqrt{k} \tan(\theta)\right)$.