If $f\colon C\to C'$ is a chain equivalence of finite chain complexes over a ring $R$, then there is a well-defined (reduced) torsion $\tau(f)\in\widetilde{K}_1(R)$. My question involves the reduced $K$-theory, i.e. why reduced?
In Davis-Kirk, they write "the reason that we use $\widetilde{K}_1$ rather than $K_1$ is that it is both messy and unnecessary for us to fuss with ordered bases." I also found that $$\tau(f\oplus f'\colon C\oplus C'\to D\oplus D')=\tau(f) + \tau(f')$$ in $\widetilde{K}_1(R)$, but that in $K_1(R)$ the sum does not split so nicely (i.e. it is $\tau(f)+\tau(f')+$ some other term involving $\tau(-1\colon R\to R)$).
How, if at all, are the ordered basis issue and the (non)-splittability of $\tau(f\oplus f')$ related?
Many mathematicians have investigated "absolute torsion" which lies in $K_1(R)$ rather than $\widetilde{K}_1(R)$. For an example of a paper on the subject, see here. You might search to find other papers.
Apparently it is not possible to find a lift of the torsion to $K_1(R)$ which has expected properties, like $\tau(gf) = \tau g + \tau f$. Hence it is messy. It is unnecessary, since in geometric applications — see 11.4, 11.5, and 11.6 of Davis-Kirk — a basis element is given by choosing an orientation on a cell in the base space and then choosing a lift of a cell in a covering space. There is no natural orientation and no natural ordering of $k$-cells, so there is really no way to define an element in $K_1(\mathbb{Z}\pi)$ determined by the geometry. But our applications do not require these arbitrary choices. Hence it is unnecessary for us to fuss with ordered bases.