 Solution

If $m_1,\dots ,m_n$ generate the module $M$, define the R-module homomorphismThis 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. ThusThis 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 $\mathbb{Z}$.