What do the finitely generated free $R$-modules look like? Let $R$ be a ring, maybe commutative or unital - but I do not know which condition is required. 
Then what are the finitely generated free $R$ modules?
Does this imply $R \simeq \bigoplus_{i \in I} R$  where $I$ is a fintie set? 
 A: With identity, everything is fine.
You haven't told us your definition of "free," but whatever it is, you will probably not have much trouble proving that every free $R$ module $F$ has an isomorphism $F\cong\bigoplus_{i\in I}R$ for some index set $I$.
If additionally $F$ is finitely generated, it is easy to conclude the index set $I$ must be finite. For, if $x_1,\ldots, x_n$ generates $F$, each one is nonzero only on a finite subset of $I$. If you take the entire collection of indices upon which $x_1,\ldots, x_n$ are nonzero, you still have a finite set. Clearly, the entire span of the $x_i$'s is supported only within the finite set, and has no nonzero entries outside of the finite set. Therefore it can only be that $I$ is finite, in that case.
A: Provided, the ring $R$ has identity (but not necessarily commutative), the free module generated by any set $I$ is isomorphic to $R^{\oplus I}$, with $i\in I$ represented as the $I$-tuple having $i$th coordinate $1$, and the rest are $0$.
A way to verify this is to prove that the universal property holds:
Let $M$ be any left $R$-module and let a mapping $f:I\to M$ be given. 
Then, for an element $(r_i)_{i\in I}\, \in R^{\oplus I}$, as we want a module homomorphism, we have to assign
$\sum_i r_i\cdot f(i)$. 
This proves the uniqueness, and also gives the formula for the existence of a homomorphism $R^{\oplus I}\to M$. 
