Let $R$ be a ring (not necessarily commutative) with multiplicative identity.
A $R$-module $M$ is called free if $M$ has a linearly independent generating set $\beta\subseteq M$.
That is, for any $m\in M$, $$m=\sum_{i=1}^{n} r_i b_i$$ for some $n\in \Bbb{N}^+$ and $r_1, r_2,...,r_n\in R$ and $b_1, b_2,...,b_n\in \beta$ and if $\sum_{i=1}^{n} r_i b_i=0$, where $b_1, b_2,..., b_n$ are distinct, then $r_1=r_2=\cdots =r_n=0$.
I can't feel the sense of the "free".
My guess is that the coefficients in the linear combination of the basis elements is "arbitrary". But it seems not to be a satisfactory explanation. Because in the non-free $\Bbb{Z}$-module $\Bbb{Z}_2\oplus \Bbb{Z}_3$, the coefficients of the generating set $\{(1,0),(0,1)\}$ is also arbitrary.