I want sums of integers $\sum_{i=1}^n c_i z_i, z_i \in \Bbb{Z}$ with coefficients in $\{0, 1\}$. Do I just state that all coefficients are always $0$ or $1$, or do I work with $\Bbb{Z}[\Bbb{Z}/2\Bbb{Z}]$, or do I work with $\Bbb{Z}/(2)[\Bbb{Z}]$?
In particular I'd still like to interpret the summations as elements of $\Bbb{Z}$ that do obviously cover $\Bbb{Z}$. Any way to do that?
Homology functors work with a coefficient group. Presumably when you speak of $0,1$-coefficients, you mean that you're working with the (unique) two-element group, for whose elements we often use the names $0$ and $1$, but which could be called, for instance, $e$ and $b$.
When you speak of "sums of integers with coefficients in $\{0, 1\}$, it's not clear whether each $z_i$ is supposed to represent a $0$-simplex in the 0-chain group of the discrete space that is of the set of integers, or whether it's supposed to be an actual number computed by summing up something like $0 \cdot 3 + 1 \cdot 5 + 1 \cdot 9 = 14$. If it's the latter, then you've apparently decided to define a pairing of elements of your two-element group $G$ with elements of the integers, $\Bbb Z$, one defined by $$ K:G \times \Bbb Z \to \Bbb Z : \begin{cases} (e, n) \mapsto 0 \\ (b, n) \mapsto n \end{cases} $$ This pairing is unfortunate, in the sense that it fails to respect the group structure. If we denote by $f_a$ the map $n \mapsto K(a, n)$, then $a \mapsto f_a$ is a map from $G$ into the homomorphism of $\Bbb Z$ to itself, which is another group $G'$. Unfortunately the map $a \mapsto f_a$ is not a homomorphism from $G$ to $G'$. In short, in doing this, you're doing something weird that doesn't respect the group structures you should be caring about.
Based on the title, I suspect that the "sum" you've written down is really a "formal sum" of items in $\Bbb Z$, which is to say a $0$-chain in the topological space $\Bbb Z$ (forget the group structure), with $G$ as your coefficient group. But it's hard to say.