I just looked at the argument, and it can be rephrased as follows:
There is a homomorphism $0\text{-cycles} \to G$ given by taking the sum of the coefficients.
If we apply this map to a boundary, we get $0$ (pretty obviously). If we apply it to $a[m]$, we get $a$. Applying this homomorphism to either side of the equation $a[m] = \text{ a boundary},$ we get $a = 0$, as claimed.
To make this answer more self-contained for others who might read it:
The degree map shows that $0\text{-cycles}/\text{boundaries}$ admits a surjection onto $G$. On the other hand, anything in the kernel is of the form $\sum_{i} a_i[m_i]$ for some elements $a_i \in G$ and points $m_i$, with $\sum_i a_i = 0.$ Thus, if we fix a point $m$, we find that $\sum_i a_i[m_i] = \sum_i a_i([m_i] - [m]) = \sum_i a_i \partial[m_i,m],$ and thus every point in the kernel of the degree map is a boundary. (Here $[m_i,m]$ is some path joining $m_i$ to $m$.) So in fact $0\text{-cycles}/\text{boundaries} \cong G.$