I could not quite get why we have parameterized r(t) as in the picture below.
In fact, parameterizing curves whenever I try to solve a problem remains to be the only step I'm having trouble with in line integrals even after reading many things on the topic and thinking that I've understood them. It makes sense after calculating, but it seems pretty arbitrary and without any well-defined method at the beginning.
Points given on the curve: $r_0 = (0,0), r_1 = (3,3a), r_2 = (2,4a)$
It is asking you to choose a parametrization for the curve between points $r_0$ and $r_1$ and then between $r_1$ and $r_2$.
It is not that the only curve between the first two points is a straight line. It could be any curve, for example, a parabola $y = \frac{a}{3}x^2$ with parametrization $(t, \frac{at^2}{3}), 0 \leq t \leq 3$. I do not think you have written down the original question but as it says mountain hike, I am assuming there is something in the question why solution has chosen a straight line. Note the line integral may vary depending on which curve you choose as the vector field is not conservative.
Now to your point on confusion in parametrization, see if the below helps in how to parametrize a straight line,
The first step to parametrize a straight line is to subtract position vector of start point from the end point so you get direction vector from start point to end point. Then the equation of oriented line is
$(0,0) + (3-0,3a-0)t = (3t, 3at)$. Now as you can see substituting $t = 0$ gives you the start point $(0,0)$ and $t = 1$ gives you the end point. So $0 \leq t \leq 1$. If we parametrized as $(t, at)$, then $0 \leq t \leq 3$.
$x'_t = 3, y'_t = 3a$ for parametrization $(3t, 3at)$ (which is nothing but the directional vector we found above by subtracting start point from the end point).
From here, the line integral follows with either the limits $0 \leq t \leq 1$ or $0 \leq t \leq 3$ depending on which parametrization we chose. Let me know if you have any questions.