I think it is slightly misleading to compare the additive and multiplicative Haar measures on local fields, although it is possible. Less tangibly, but more indicative of the nature of the situation, is the fact that the multiplicative group of a non-archimedean local field has a unique maximal compact (and open) subgroup, the local units, which (correctly) suggests the normalization of the Haar measure, to give the local units measure 1. In contrast, the additive group of a non-archimedean local field is an ascending union of compact open subgroups.
The story of the measure on the compact A/k is perhaps the story that the absolutely ramified non-archimedean places have their additive measure normalized in a certain way depending on the local "different", and because the global different is the product of the local, and the discriminant-squared is the ideal-norm of the different, amazingly the inherited measure on the quotient A/k is 1.
Rather than choose fundamental domains, the measure on a quotient $G/H$ of abelian topological groups is completely determined by the measure on $G$ and that on $H$, by $\int_G f = \int_{G/H} \int_H f(gh)\,dh\;dg$ for compactly supported continuous $f$.
The measure on the ideles is determined by the local measures everywhere, which at finite places give the local units measure 1. Then $J/k^\times$ has a uniquely-determined measure determined by counting measure on $k^\times$ and the measure on $J$. Then $(J/k^\times)/(J^1/k^\times)\approx (0,+\infty)$ is given the usual measure $dx/x$ via that natural isomorphism. By the relation of measures, as above, this uniquely specifies the measure on $J^1/k^\times$.
It is a further exercise to show that with this canonically defined measure the total measure of $J^1/k^\times$ is the usual $2^{r_1}(2\pi)^{r_2}hR/D^{1/2}w$.
