EDIT 1/5/2010: I was a little dissatisfied with the informality of what I wrote below, so here is a somewhat more formal writeup. The statement is somehow geometrically obvious, but the proof is still a bit nice.
EDIT: I initially claimed the sequence below is short exact, which is false; it is right exact. It is fixed below, with some explanation, and a bit of a geometric explanation of what's going on.
I know little about the theory of linear independence over $mathbb{Q}$, but I'll attempt an answer to this part of the question:
Are there relationships between the (co)homology groups of the covering and the residues?
The answer is "yes." Let $f$ be a meromorphic function on $mathbb{C}$, and for convenience let's assume that it has poles $z_1, ..., z_n$ of order $1$, and no other singularities. Let $g$ be an antiderivative of $f$. Then the Riemann surface of $f$ is $M_f=mathbb{C}-{z_1, ..., z_n}$ and the Riemann surface of $g$, which we will denote by $M_g$, is a covering space of $M_f$ with covering map $pi: M_gto M_f$. Let $Vsubset mathbb{C}$ be the $mathbb{Q}$-vector space spanned by $operatorname{Res}_{z_1}(f), ..., operatorname{Res}_{z_n}(f)$. Then there is a short exact sequence $$H_1(M_g, mathbb{Q})overset{H_1(pi)}{longrightarrow} H_1(M_f, mathbb{Q})overset{int}{longrightarrow} Vto 0,$$
where the map $int$ is given as follows. Namely, $int: H_1(M_f, mathbb{Q})to V$ is given by $[gamma]mapsto frac{1}{2pi i}int_gamma f~operatorname{dz}$.
Let's elucidate the connection to the linear independence of $operatorname{Res}_{z_1}(f), ..., operatorname{Res}_{z_n}(f)$ over $mathbb{Q}$. $H_1(M_f, mathbb{Q})$ is a $mathbb{Q}$-vector space with a basis of cycles $[lambda_1], ..., [lambda_n]$ corresponding to the punctures $z_1, ..., z_n$. Then the map $int$ sends $[lambda_i]$ to $operatorname{Res}_{z_i}(f)$. So the image of $H_1(M_g, mathbb{Q})$ in $H_1(M_f, mathbb{Q})$ is precisely the vector space of relations between the residues of $f$.
Added: We can extend this right exact sequence into a longer sequence. In particular, by covering space theory we have that $pi_1(M_g)to pi_1(M_f)$ is an injection. It is easy to see that the commutator subgroup of $pi_1(M_f)$ is contained in the image of $pi_1(M_g)$. By the Hurewicz theorem $$H_1(M_g, mathbb{Q})simeq pi_1(M_g)^{Ab}underset{mathbb{Z}}{otimes} mathbb{Q}.$$
So the kernel of the map $H_1(M_g, mathbb{Q})to H_1(M_f, mathbb{Q})$ is given by the image of $[pi_1(M_f), pi_1(M_f)]$ (which is contained in $pi_1(M_g)$) in $H_1(M_g, mathbb{Q})$. One can extend the exact sequence further back by looking at quotients of commutators in this manner.
This first extension has a geometrical interpretation. Namely, let $h$ be a meromorphic function whose poles have the same locations as those of $f$, but whose residues are linearly independent over $mathbb{Q}$. Then the antiderivative of $h$, denoted $s$ has Riemann surface $M_s$, which is a covering space over $M_f$, with covering map $pi': M_sto M_f$. By the properties of covering spaces, $pi'$ factors through $pi$, and it is not hard to see that $pi_1(M_s)$ is exactly the commutator subgroup of $pi_1(M_f)$. Then the sequence $$H_1(M_s, mathbb{Q})overset{H_1(pi')}{longrightarrow} H_1(M_g, mathbb{Q})overset{H_1(pi)}{longrightarrow} H_1(M_f, mathbb{Q})overset{int}{longrightarrow} Vto 0,$$
is exact, and coincides with the sequence described above.
I don't know if the continuing left extensions of this sequence have similar geometric interpretations. Also, it would be nice to have a naturally arising description of this sequence, rather than the somewhat ad hoc one I've given.
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