[proofplan] Choose a finite generating family for $M$ over $R$ and localize each generator by putting denominator $1$. We show that these localized elements generate every fraction in $S^{-1}M$. The key calculation is that if $m = \sum_i r_i m_i$ in $M$, then the fraction $m/s$ decomposes over $S^{-1}R$ as $\sum_i (r_i/s)(m_i/1)$. [/proofplan] [step:Choose localized candidates from a finite generating family] Since $M$ is finitely generated as an $R$-module, there exist an integer $k \ge 0$ and elements $m_1, \ldots, m_k \in M$ such that every element of $M$ is an $R$-linear combination of $m_1, \ldots, m_k$. Define elements $u_1, \ldots, u_k \in S^{-1}M$ by \begin{align*} u_i := \frac{m_i}{1} \end{align*} for each $i \in \{1, \ldots, k\}$. We will prove that $u_1, \ldots, u_k$ generate $S^{-1}M$ as an $S^{-1}R$-module. [/step] [step:Rewrite an arbitrary localized element using the localized generators] Let $x \in S^{-1}M$. By the construction of module localization, there exist $m \in M$ and $s \in S$ such that \begin{align*} x = \frac{m}{s}. \end{align*} Since $m_1, \ldots, m_k$ generate $M$ over $R$, there exist elements $r_1, \ldots, r_k \in R$ such that \begin{align*} m = \sum_{i=1}^{k} r_i m_i. \end{align*} For each $i \in \{1, \ldots, k\}$, define $a_i \in S^{-1}R$ by \begin{align*} a_i := \frac{r_i}{s}. \end{align*} Using the $S^{-1}R$-module structure on $S^{-1}M$, we compute \begin{align*} \sum_{i=1}^{k} a_i u_i = \sum_{i=1}^{k} \frac{r_i}{s}\cdot \frac{m_i}{1}. \end{align*} By the defining multiplication rule for localized modules, \begin{align*} \sum_{i=1}^{k} \frac{r_i}{s}\cdot \frac{m_i}{1} = \sum_{i=1}^{k} \frac{r_i m_i}{s}. \end{align*} By the defining addition rule for fractions with common denominator in $S^{-1}M$, \begin{align*} \sum_{i=1}^{k} \frac{r_i m_i}{s} = \frac{\sum_{i=1}^{k} r_i m_i}{s}. \end{align*} Therefore \begin{align*} \sum_{i=1}^{k} a_i u_i = \frac{m}{s} = x. \end{align*} [guided] Take an arbitrary element $x \in S^{-1}M$. The point of localization is that every element of $S^{-1}M$ is represented by a fraction with numerator in $M$ and denominator in $S$, so there exist $m \in M$ and $s \in S$ such that \begin{align*} x = \frac{m}{s}. \end{align*} Now use the only hypothesis on $M$: the elements $m_1, \ldots, m_k$ generate $M$ as an $R$-module. Thus there are coefficients $r_1, \ldots, r_k \in R$ satisfying \begin{align*} m = \sum_{i=1}^{k} r_i m_i. \end{align*} We want to express $x$ as an $S^{-1}R$-linear combination of the localized generators $u_i = m_i/1$. The natural coefficients are the localized ring elements $r_i/s$, so for each $i \in \{1, \ldots, k\}$ define \begin{align*} a_i := \frac{r_i}{s} \in S^{-1}R. \end{align*} Then the localized module action gives \begin{align*} a_i u_i = \frac{r_i}{s}\cdot \frac{m_i}{1} = \frac{r_i m_i}{s}. \end{align*} Summing these identities in $S^{-1}M$ and using the addition rule for fractions with the same denominator gives \begin{align*} \sum_{i=1}^{k} a_i u_i = \sum_{i=1}^{k} \frac{r_i m_i}{s}. \end{align*} The common-denominator addition rule in the localized module gives \begin{align*} \sum_{i=1}^{k} \frac{r_i m_i}{s} = \frac{\sum_{i=1}^{k} r_i m_i}{s}. \end{align*} Substituting the chosen expression for $m$ yields \begin{align*} \frac{\sum_{i=1}^{k} r_i m_i}{s} = \frac{m}{s} = x. \end{align*} Thus the arbitrary localized element $x$ is an $S^{-1}R$-linear combination of $u_1, \ldots, u_k$. [/guided] [/step] [step:Conclude finite generation over the localized ring] The preceding step shows that every element $x \in S^{-1}M$ lies in the $S^{-1}R$-submodule generated by $u_1, \ldots, u_k$. Hence $S^{-1}M$ is generated by the finite family \begin{align*} \frac{m_1}{1}, \ldots, \frac{m_k}{1} \end{align*} as an $S^{-1}R$-module. Therefore $S^{-1}M$ is finitely generated over $S^{-1}R$. [/step]