= Solution
A subset $S\subseteq M$ freely generates $M$ when every $m\in M$ has a unique expression
$$
m=\sum_{s\in S}r_ss
$$
with $r_s\in R$ and only finitely many nonzero coefficients. Equivalently, $S$ is a <basis of a module> and $M$ is the <free module> on $S$.
If $S$ freely generates $M$ and $f:S\to N$ is any function, define
$$
\phi\left(\sum_sr_ss\right)=\sum_sr_sf(s).
$$
Unique coordinates make $\phi$ well defined; it is an $R$-module homomorphism and is the only extension of $f$. Conversely, apply the proposed universal property to the free module $F=R^{(S)}$ and the inclusion of $S$ into $M$. It gives maps $F\to M$ and $M\to F$ extending the corresponding functions on $S$. Uniqueness makes both composites identity maps, so $M\cong F$ and $S$ freely generates $M$. This is the <universal property of a free module>.
Now let $T$ generate the free module $M$ and $|T|=m$. Choose a <maximal ideal> $\mathfrak m$ of the nontrivial ring $R$. Then $k=R/\mathfrak m$ is a field, and
$$
M/\mathfrak mM
$$
is a $k$-vector space. The images of a basis $S$ of $M$ form a vector-space basis, while the images of $T$ span it. Therefore
$$
|S|=\dim_k(M/\mathfrak mM)\leq m.
$$
In particular $S$ is finite. Applying this result in both directions to two finite bases shows that they have equal cardinality, the <rank of a free module> $\operatorname{rk}M$.
A Euclidean domain is a <principal ideal domain>. By the <submodule theorem for free modules over a principal ideal domain>, every submodule $N$ of the finite-rank free module $M$ is free. Since a basis of $M$ generates it, the preceding inequality applied in the standard proof gives
$$
\boxed{\operatorname{rk}N\leq\operatorname{rk}M}.
$$
The <primary decomposition theorem for finitely generated modules over a principal ideal domain> states that
$$
M\cong R^r\oplus
\bigoplus_{p}\bigoplus_j R/(p^{\,e_{p,j}}),
$$
where $p$ ranges over finitely many nonassociate irreducibles and the positive exponents are uniquely determined up to order. Equivalently, the torsion part decomposes into its primary cyclic summands.
Let $H$ be a finite subgroup of the multiplicative group of a field. It is a finite abelian group, hence the theorem over $\mathbb Z$ gives an invariant-factor decomposition
$$
H\cong C_{d_1}\times\cdots\times C_{d_r},
\qquad d_1\mid\cdots\mid d_r.
$$
Its exponent is $d_r$, so every element of $H$ is a root of $X^{d_r}-1$. A degree-$d_r$ polynomial over a field has at most $d_r$ roots, whence
$$
|H|\leq d_r.
$$
But $d_r\leq|H|$, and equality forces all earlier factors to be trivial. Thus
$$
\boxed{H\text{ is cyclic}}.
$$
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