In looking back at my university's past topology qualifying exams, it seems that in most years there is a question of the form
Let $X=S^1\vee S^1$. Draw a picture of the cover of $X$ corresponding to the subgroup $H=\langle\text{some generators}\rangle$ of $\pi_1(X)=\langle a,b\rangle$.
When I see a solution to this type of problem, I can see why it satisfies the right conditions, typically having to with the fact that traversing a series of loops corresponding to a generator of $H$ brings you back to the basepoint, etc. However, I'm not good at coming up with these pictures myself. Is there a general method or "trick"?
Maybe I can give some explicit examples of $H$ and people could share their "thought process" for the solution they come to. For example,
(a) take $H=\langle ab,ba\rangle$
(b) take $H=\langle(a^2b)^3,(ab)^3,bab,a^2ba^{-1}\rangle$
(c) take $H=\langle aba^{-1}b^{-1},ab^{-1}a^{-1}b,a^{-1}bab^{-1},a^{-1}b^{-1}ab\rangle$
Here's a general method.
To draw the covering corresponding to $H$, you let your vertices be indexed by the cosets of $H$ in $G$; your basepoint will correspond to the coset $H$.Then you draw an $a$ edge from $C$ to $Ca$ for every coset $C$; similarly for $b$.
I can't imagine there's an easier way to handle the general case. Something helpful here is that $H$ is a free group, so it has some number $n$ of generators. Thus your covering should be homotopic to the wedge of $n$ circles. For example, that helps me realize that the covering for (a) should look like
where each tree extends infinitely (just like the Cayley graph / universal cover for $G$).