The definition for a free $R$-module is the following:
An $R-$module $F$ is free if there exists a linearly independent generating set $B=\{b_i\}_{i\in I}$, called a basis. Note that in this case $F$ is isomorphic to a direct sum of copies of $R$. The isomorphism $F \to \bigoplus_{i \in I} R_i$ sends a linear combination $\sum_{i\in I}r_ib_i$ to the tuple $(r_{i})_{i\in I}$.
Now I am trying to understand this definition. Let's say we call the isomorphism $\sigma$ and so for some element in $F$, $m=r_{1}b_{1}+r_{2}b_{2}$, this map gives us the following: $\sigma(r_{1}b_{1}+r_{2}b_{2})=(r_{1},r_{2})$. The first step is to show that $\sigma$ is injective. Thus, we need to show that if $\sigma(r_{1}b_{1}+r_{2}b_{2})=(0,0)$, it follows that $r_{1}b_{1}+r_{2}b_{2}=0$.
But I am stuck because I do not know where to start since we have one coordinate on the left-hand side of the equation and two coordinates on the right. Where do I go from here? Thank you.
The key here is linear independence. This is required for the map $r_1 b_1 + r_2 b_2 \mapsto (r_1, r_2)$ to be well-defined. How do we know we can't write $r_1 b_1 + r_2 b_2 = s_1 b_1 + s_2 b_2$ with different coefficients? Well by definition, $b_1, b_2$ are linearly independent if and only if $r_1$ and $r_2$ are unique. So we have a well-defined map. This is the key step. Injectivity is automatic since $r_1 b_1 + r_2 b_2 \mapsto (0, 0)$ if and only if $(r_1, r_2) = (0, 0)$ if and only if $r_1 b_1 + r_2 b_2 = 0b_1 + 0b_2 = 0$.