We defined the $n$-chain group as follows,
$$C_n(X) = \bigg\{ \sum_{v \in V} n_v [v_1, \dots, v_n] : n_v \in \mathbb{Z}, \hspace{2mm} n_v=0 \hspace{2mm} \text{for all but finitely many} \hspace{1mm} n_v\bigg\}$$
I am not sure why it is important that $n_v=0$ for all but finitely many coefficients.
Is this the same thing as saying that $C_n(X)$ is a formal sum, and if so, why can it not be an arbitrary sum?
One possible reason is that if we allowed infinite formal sums then it would be impossible to define the boundary map. Suppose our simplicial complex consists of all simplices of the form $[v_0,v_i]$ with $i\geq 1$. Consider the formal sum $$\sum_{i=1}^{\infty}{[v_0,v_i]}$$ The boundary of this linear combination would be $$\sum_{i=1}^{\infty}{([v_i]-[v_0])}$$ and this is not well-defined because the coefficient of $[v_0]$ would be $-\infty$.
The chain complex is defined as the free abelian group generated by the simplices, and this consists of all formal finite linear combinations of simplices.