What is the determinant of a complex of vector spaces？

The appendix A of the book that you mention is probably the best reference for determinants!

• If $V$ is a $k$-vector space of dimension $d$, then its determinant is the one-dimensional vector space
  $$\det (V) \stackrel{def}{=} \Lambda^d (V).$$
  If $V = 0$, we set
  $$\det (0) \stackrel{def}{=} k.$$

• If $V^\bullet$ is a complex of finite dimensional vector spaces such that $V^i = 0$ for almost all $i$ (all but finitely many $i$), then one defines
  $$\det (V^\bullet) \stackrel{def}{=} \bigotimes_{i\in \mathbb{Z}} \det (V^i)^{(-1)^i}.$$
  Here tensor products are over $k$, and the sign $-1$ denotes taking the dual vector space: $V^{-1} \stackrel{def}{=} V^\vee$.

• In the above situation, one can show that
  $$\tag{*} \bigotimes_{i\in \mathbb{Z}} \det (V^i)^{(-1)^i} \cong \bigotimes_{i\in \mathbb{Z}} \det (H^i (V^\bullet))^{(-1)^i},$$
  where $H^i (V^\bullet)$ denote the cohomology spaces.

• In general, if $V^\bullet$ is a complex that has finite dimensional cohomology spaces, and $H^i (V^\bullet) = 0$ for almost all $i$, then we may use the right hand side (*) as the definition:
  $$\det (V^\bullet) \stackrel{def}{=} \bigotimes_{i\in \mathbb{Z}} \det (H^i (V^\bullet))^{(-1)^i}.$$

It is worth noting that the whole point of $\det$ is that it is functorial. Namely, $V^\bullet \xrightarrow{\cong} W^\bullet$ induces an isomorphism of one-dimensional vector spaces $\det (V^\bullet) \xrightarrow{\cong} \det (W^\bullet)$.

Here’s one helpful analogy. Note that the formula (*) reminds the formula for the Euler characteristic

$$\sum_{i\in\mathbb{Z}} (-1)^i \dim V^i = \sum_{i\in\mathbb{Z}} (-1)^i \dim H^i (V^\bullet),$$
and the proof of (*) actually goes along the same lines: it is based on the fact that if we have a short exact sequence of finite dimensional vector spaces
$$0 \to V' \to V \to V'' \to 0$$
then there is a canonical isomorphism
$$\det V \cong \det V' \otimes_k \det V''$$
(this is analogous to the formula $\dim V = \dim V' + \dim V''$.)