Let $K$ be an oriented knot in $S^3$ together with a framing $n$. Let $K(a,b)$ be the oriented link obtained by taking $a$ copies of the $n$-pushoff of $K$ with the same same orientation as $K$ and $b$ copies of the $n$-pushoff of $K$ with the opposite orientation as $K$. In particular, $K(1,0) = K$ and $K(0,1)$ is the reverse of $K$. 


What can be said about the signature function $\sigma\_{K(a,b)}(\lambda)$ in terms of $\sigma\_K(\lambda)$? 


Edit: The answer is yes - in fact there is a nice formula for more general satellite constructions. In "Signatures of iterated torus knots" by Litherland, it is show that $\sigma\_{K(a,b)}(\lambda) = \sigma\_K(\lambda^{a-b})$ (See Theorem 2 and Remark 2 following it).


Can we say anything about the nullity $n\_{K(a,b)}(\lambda)$ in terms of some info about $K$? In light of the above Edit, I should probably just ask about how the nullity function behaves under satellite operations. 



Signature/nullity function for a link obtained by parallel pushoffs of a knot?
As Marco mentions, it appears you are studying cables. These are particular examples of satellite operations. As you already noticed, Litherland proves that if $S$ is a winding number $n$ satellite with pattern $P$ and companion $C$, then
$$ \sigma\_S(\omega)=\sigma\_P(\omega)+\sigma\_C(\omega^n). $$ 
For the nullity, I think the formula is similar:
$$ \eta\_S(\omega)=\eta\_P(\omega)+\eta\_C(\omega^n). $$ 
I think a proof can be found in here in a more general setting, namely "splicing":
<https://arxiv.org/pdf/1802.01836.pdf> (Theorem 5.2).
The way to recover satellites from splices is for instance explained here:
<https://arxiv.org/pdf/1409.5873.pdf> (Section 2.4).


For the nullity, you can also prove this "by hand". I think that the most painless way is to view $\eta\_K(\omega)$ as the dimension of a twisted homology vector space: $ \eta\_K(\omega)=\operatorname{dim}\_{\mathbb{C}} H\_1(S^3 \setminus K;\mathbb{C}^\omega)$. You can then run a Mayer-Vietoris argument for the satellite knot exterior and the result will follow.


