Let $A$ be a ring. We say an $A$-module $M$ is a generator if for every $A$-module $N$ there is a surjective $A$-homomorphism $M^{(I)} \rightarrow N$ where $M^{(I)}$ denotes the $|I|$ copies of the direct sum of $M$. Why is it true that every free $A$-module $T$ is a generator for $A$-mod?
Suppose $T$ is a free $A$-module then $T$ has a basis, say $\{t_{i}: i \in I\}$ and let $N$ be an arbitrary $A$-module. How do we define a surjective map from this basis to an arbitrary $A$-module $N$?
That $T$ is a free $A$-module is equivalent to say that $T\cong A^{(I)}$ for some nonempty set $I$ (it is the direct sum of each $At_i\ (\cong A)$ if you start out from a basis $(t_i)_{i\in I}$).
To be a generator, as you cited, means that, for all modules $N$ there is a set $J$ and a surjective homomorphism $T^{(J)}\to N$, so it is not a $T\to N$ surjection in general, which you now started to look for.
Of course, $A$ itself is a generator: if a module $N$ is generated by a set $(x_i)_{i\in J}$, then it gives a surjective homomorphism $A^{(J)}\to N$.
As $T=A^{(I)}$, it is all the more a generator: for the same $N$ with generators $(x_i)_{i\in J}$ the same $J$ will work, because $|I|\cdot|J|\ge |J|$, so we have surjections $T^{(J)}\to A^{(J)}\to N$.
