Here's a graph of the line $y = 2x + 3$ along with some rays outward from the origin.
Each ray is labeled with the angle of the ray in polar coordinates.
The rays labeled $\theta_1$ and $\theta_2$ lie along the line through the origin parallel to the line $y = 2x + 3$. This line has slope $2$ and the angle of the
ray $\theta_1$ is therefore $\theta_1 = \arctan(2).$
The ray $\theta_2$ is in the exact opposite direction from $\theta_1.$
To rotate the ray $\theta_1$ counterclockwise onto $\theta_2,$
we have to rotate it $180$ degrees, which is $\pi$ radians.
Therefore $\theta_2 = \theta_1 + \pi = \arctan(2) + \pi.$
As we rotate a ray so that its angle increases from $\theta_1$ to $\theta_2,$
at each angle (for example $\theta_3,$ $\theta_4,$ and $\theta_5$),
the ray intersects the line and we can use that angle $\theta$
and the radius $r = f(\theta) = 3 / (\sin \theta - 2 \cos \theta)$
to plot one of the points on the line.
As the ray sweeps through all the angles between $\theta_1$ and $\theta_2,$
the polar coordinates $(f(\theta), \theta)$ plot every point on the line.
We want all of those angles to be in the domain of $f$ so that we can plot the whole line.
If you stopped at some angle before $\arctan(2) + \pi,$ for example $\theta_4,$ you would only have plotted part of the line.
All the parts of the line at angles between $\theta_4$ and $\arctan(2) + \pi$
would be missing.
Note that $\arctan(2) + \pi$ actually is not "allowed,"
because the upper bound of the domain is described by
$\theta < \arctan(2) + \pi$; notice how it is $<$ rather than $\leq,$
which tells us that the angle $\theta$ can never exactly equal $\arctan(2) + \pi.$
We cannot include either $\arctan(2)$ or $\arctan(2) + \pi$ in the domain,
because the rays at those angles are parallel to the line and never intersect it,
so there is no value we can set $r = f(\theta)$ to in order for
$(f(\theta), \theta)$ to be polar coordinates of a point on the line.