Let $X$ be a Noetherian scheme and $\mathcal{G} \in \text{Coh}(X)$. Hartshorne III.6.3(c) states
$$ \text{Ext}^i(\mathcal{O}_X, \mathcal{G}) \cong H^i(X, \mathcal{G}) $$
as modules. What does the $\mathbb{Z}$-module structure on $H^i(X, \mathcal{G})$ correspond to in the derived category $\textbf{D}^b(\text{Coh}(X))$? More precisely, recall that a morphism in $\textbf{D}^b(\text{Coh}X)$ of $\mathcal{O}_X$ to $\mathcal{G}$ is given by a hammock diagram (which I do not want to type out for lack of tikz). Such hammocks represent cohomology classes from the sheaf $\mathcal{G}$. Since the cohomology group is in fact a group, what are the operations on the level of hammock diagrams which represent addition/inverse of elements in the cohomology group?
If I understand you correctly, you want to know why the derived category is additive, and what represents the sum $f+g$ of two morphisms. (Note that the derived category of any abelian category is additive, not just the derived category of coherent sheaves).
In terms of zig-zag, you have the following description of $f+g$ : $f$ and $g$ are represented by zig-zags of morphisms of complexes (where left arrows are quasi-isomorphisms) $$f:A\overset{f_1}{\underset{qis}\longleftarrow}T_1\overset{f_2}{\longrightarrow}T_2\overset{f_3}{\underset{qis}\longleftarrow}T_3\longrightarrow ...\overset{f_n}{\underset{qis}\longleftarrow}T_n\overset{f_{n+1}}\longrightarrow B $$ $$g:A\overset{g_1}{\underset{qis}\longleftarrow}S_1\overset{g_2}{\longrightarrow}S_2\overset{g_3}{\underset{qis}\longleftarrow}S_3\longrightarrow ...\overset{g_m}{\underset{qis}\longleftarrow}S_m\overset{f_{m+1}}\longrightarrow B.$$ You can assume that $n=m$ (by adding identities if one zig-zag is too short), so taking direct sum, you have a zig-zag : $$f\oplus g:A\oplus A\overset{f_1\oplus g_1}{\underset{qis}\longleftarrow}T_1\oplus S_1\overset{f_2\oplus g_2}{\longrightarrow}T_2\oplus S_2\longleftarrow ...\overset{f_n\oplus g_n}{\underset{qis}\longleftarrow}T_n\oplus S_n\overset{f_{n+1}\oplus g_{n+1}}\longrightarrow B\oplus B $$
The sum $f+g$ is then the composition $A\rightarrow A\oplus A\overset{zig-zag}\longrightarrow B\oplus B\rightarrow B$ where the arrow on the left is the diagonal and the arrow on the right is the sum. In fact, you see that the structure of zig-zag is really not important. The sum of two morphisms is just given by the composite above, where $zig-zag$ is the diagonal matrix $\pmatrix{f&0\\0&g}$ in the derived category.
As a consequence, you can simply replace $B$ by an injective resolution $I^.$. The sum $f+g$ is then the sum of the two corresponding morphisms $A\rightarrow I^.$
For the opposite, just replace $f_1$ (or any of the $f_i$'s) by $-f_1$ (or $-f_i$). You can also compose on the left or on the right by $-id$.