I am trying to find a proof (or counterexample) of the following claim: the Alexander polynomial of the 2-cable of a knot (where the blackboard framing is considered) is zero. That is, the Alexander polynomial of the 2-component link obtained by doubling the strand in the obvious way is zero.
My gut feeling says this shouldn't be correct but (if I am not making any mistake) this needs to be true because of the fact that the Alexander polynomial of the Whitehead double is 1. Considering the skein relation for Alexander of the Whitehead double of a knot (say $L_+$) we obtain that $L_-$ is isotopic to the unknot and $L_0$ to the 2-cable of the knot. For $\Delta_{Wh(K)}$ to be 1, $\Delta_{2-cable(K)}$ needs to be 0.
One definition of the Alexander polynomial of an oriented link $L$ is that you take a Seifert surface $\Sigma$, which is a connected oriented surface whose boundary is $L$ (with the oriented boundary matching the orientations of $L$) and whose interior is disjoint from $L$, then calculating the Seifert form $\alpha:H_1(\Sigma)\times H_1(\Sigma)\to\mathbb{Z}$, where $\alpha([a],[b])=\operatorname{lk}(a,b^+)$ for simple closed curves $a,b\subset \Sigma$ (the curve $b^+$ is from pushing the loop off the surface in the direction of surface normal of $\Sigma$), then letting $A$ be a matrix for $\alpha$ with respect to some basis, and finally computing $\det(tA-A^T)$.
For a 2-cabling $L$ of a knot $K$, we need to decide on how we orient the components. The usual choice is to have both components be co-oriented, but one could potentially choose opposite orientations. These give different answers.
Let's consider the oppositely oriented case first. There is a natural choice of Seifert surface $\Sigma$, which is the annulus that runs along with the two components, parallel to the blackboard for the blackboard framing -- note that the oriented boundary of the annulus is indeed giving the components of $L$ opposite orientations. Thus, $H_1(\Sigma)=\mathbb{Z}$. The core of the annulus $a\subset\Sigma$ (with some orientation) generates $H_1(\Sigma)$. If $L_1$ and $L_2$ are the two components, what we can do is isotope $a$ to $L_1$ and, away from $a$, isotope the pushoff $a^+$ to $L_2$. Thus, $\operatorname{lk}(a,a^+)=\operatorname{lk}(L_1,L_2)$, which is the writhe $w(K)$ of the blackboard framed diagram. Thus, the matrix of the Seifert form can be given as the $1\times 1$ matrix $[w(K)]$ with respect to the $\{a\}$ basis. We calculate that $$ \Delta_L(t) = \det(tA-A^T) = \det(t[w(K)]-[w(k)]^T)=w(k)(t-1). $$ (And remember this is only defined up to multiplication by $\pm t^k$.) So we see that the Alexander polynomial is not telling you very much about the link.
The case where the components are co-oriented is more involved. I point you to Theorem 6.15 of Lickorish's "An Introduction to Knot Theory" (p.60), though as it is stated it is for the case where you make a satellite knot rather than a link, it might need some modification. The main idea is that if you have a Seifert surface for $K$, then you can take two parallel copies $\Sigma_0$ and $\Sigma_1$ (from pushing a copy of the Seifert surface off itself), and then you can add half-twisted strips to the boundaries to get a connected surface $\Sigma$ whose boundary is $L$. One can split up $H_1(\Sigma)$ into $H_1(\Sigma_0)\oplus H_1(\Sigma_1)\oplus H_1(\Sigma')$ where $\Sigma'$ consists of the strips along with a tubular neighborhood of the boundaries of the surfaces. Then it should be possible to work out a Seifert matrix. Sorry to leave it here -- I did calculate some Alexander polynomials of cablings to see that they do not tend to be trivial. The most basic example is $(2,2n)$ torus links (i.e., 2-cablings of the unknot), whose Alexander polynomials are given by $(1-t^{2n})/(1+t)$.
If it turns out to be fine applying the theorem to this case, then the theorem implies that $$\Delta_L(t) = \Delta_T(t)\Delta_K(t^2) = \frac{1-t^{2w(K)}}{1+t}\Delta_K(t^2),$$ where $T$ is the torus link $T(2,2w(K))$ (with both components co-oriented).