Because the trace is an additive invariant of a representation. If $V= \bigoplus a_iV_i$ is a $G$-module, then for each $g\in G$ $$\chi_V(g) = Tr(\pi(g)) = Tr(\sum_i a_i\pi_i(g)) = \sum a_i Tr(\pi_i(g)) = \sum a_i\chi_i(g). $$
In view of the comments below, let me supplant my answer with the rest of the story (I am working over $\mathbb{C}$ from now on).
The character of a $G$-representation is the class function associated to $\pi:G\curvearrowright V$ given by $\chi_V(g) = Tr(\pi(g))$. We saw above that a non-negative integer linear combination of the $\chi_i$ associated to the irreducible representations is a character.
Conversely, every character is a non-negative integer linear combination of the $\chi_i$. To show this: let $V$ be a $G$-space. By complete reducibility, $V$ is a direct sum of irreducibles. Gathering multiplicities, $V\simeq \oplus a_i V_i$ where $a_i$ are the multiplicities, possibly zero, and $V_i$ runs through irreps. Then the same computation as above shows that $\chi_V$ is a non-negative integer linear combination of the $\chi_i$. The $\chi_i$ are independent because they are orthogonal with respect to the usual inner product, so the expression of $\chi_V$ is unique.
This shows that the set $C$ of characters of the various complex representations of $G$ are precisely the non-negative cone in the integer lattice spanned by the $\chi_i$: $$C=\bigoplus_i \mathbb{Z}_{\geq 0}\chi_i.$$