= Solution
If $m_1,\ldots,m_n$ generate the <module> $M$, define the <R-module homomorphism>
$$
T:R^n\longrightarrow M,\qquad (r_1,\ldots,r_n)\longmapsto\sum_{j=1}^nr_jm_j.
$$
This is a <surjection> by the definition of generation. Conversely, an <R-module homomorphism> that is a <surjection> sends the standard <basis> vectors of the <free module> $R^n$ to a generating family, since every vector in $R^n$ is their linear combination. Thus
$$
\boxed{M\text{ is finitely generated}\iff M\text{ is a quotient of some finite free module }R^n.}
$$
This is the <finite generation as a quotient of a finite free module> criterion. The zero <module> causes no exception: it is generated by the empty family, or by one zero element.
For the requested example, take $M=(\mathbb Z/2\mathbb Z)^2$, with the usual integer scalar action. Reducing both coordinates modulo two gives an <R-module homomorphism> that is a <surjection> $\mathbb Z^2\to M$ with nonzero <kernel> $2\mathbb Z^2$, so it is not an <isomorphism>. Every image of $\mathbb Z$ is a <cyclic module>, whereas every element of $M$ has additive order at most two and $M$ has four elements. Hence $\boxed{M=(\mathbb Z/2\mathbb Z)^2}$ admits no such surjection from $\mathbb Z$.
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