[proofplan]
Choose finitely many $R$-module generators $m_1,\ldots,m_k$ of $M$. We prove that the corresponding tensors $1_S \otimes m_1,\ldots,1_S \otimes m_k$ generate $S \otimes_R M$ as an $S$-module. Since every element of a [tensor product](/page/Tensor%20Product) is a finite sum of pure tensors, it is enough to rewrite an arbitrary pure tensor $s \otimes m$ using the $R$-linear expansion of $m$ in terms of the $m_i$. The balancing relation in the tensor product converts each $R$-coefficient into multiplication by its image in $S$, giving an $S$-linear combination of the chosen tensors.
[/proofplan]
[step:Choose finite generators and the induced $S$-module structure]
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 $m \in M$ can be written in the form
\begin{align*}
m = \sum_{i=1}^{k} r_i m_i
\end{align*}
for some $r_1,\ldots,r_k \in R$.
The $R$-module structure on $S$ is given by
\begin{align*}
r \cdot s = \varphi(r)s
\end{align*}
for $r \in R$ and $s \in S$. The standard $S$-module structure on $S \otimes_R M$ is defined on pure tensors by
\begin{align*}
a \cdot (s \otimes m) = (as) \otimes m
\end{align*}
for $a,s \in S$ and $m \in M$, and then extended additively.
Define the finite subset $G \subset S \otimes_R M$ by
\begin{align*}
G = \{1_S \otimes m_i : 1 \le i \le k\}.
\end{align*}
If $k=0$, this set is empty, and the displayed generating condition says exactly that $M=0$. In that case every pure tensor $s \otimes m$ is zero, so $S \otimes_R M=0$, which is generated by the empty set as an $S$-module. Hence we may argue uniformly with the same notation.
[/step]
[step:Rewrite each pure tensor using the chosen generators]
Let $s \in S$ and $m \in M$. Choose $r_1,\ldots,r_k \in R$ such that
\begin{align*}
m = \sum_{i=1}^{k} r_i m_i.
\end{align*}
Using additivity of the tensor product in the second variable, we obtain
\begin{align*}
s \otimes m = \sum_{i=1}^{k} s \otimes (r_i m_i).
\end{align*}
Using the balancing relation for the tensor product over $R$, with $S$ viewed as an $R$-module through $\varphi$, each summand satisfies
\begin{align*}
s \otimes (r_i m_i) = (\varphi(r_i)s) \otimes m_i.
\end{align*}
By the definition of the $S$-module structure on $S \otimes_R M$,
\begin{align*}
(\varphi(r_i)s) \otimes m_i = (\varphi(r_i)s) \cdot (1_S \otimes m_i).
\end{align*}
Therefore
\begin{align*}
s \otimes m = \sum_{i=1}^{k} (\varphi(r_i)s) \cdot (1_S \otimes m_i).
\end{align*}
Thus every pure tensor lies in the $S$-submodule generated by $G$.
[guided]
The goal is to show that the tensors $1_S \otimes m_i$ generate the whole scalar extension. So take a typical pure tensor $s \otimes m$, where $s \in S$ and $m \in M$. Because $m_1,\ldots,m_k$ generate $M$ as an $R$-module, there are elements $r_1,\ldots,r_k \in R$ such that
\begin{align*}
m = \sum_{i=1}^{k} r_i m_i.
\end{align*}
Now substitute this expression into the second tensor factor. The tensor product is additive in each variable, so
\begin{align*}
s \otimes m = s \otimes \sum_{i=1}^{k} r_i m_i = \sum_{i=1}^{k} s \otimes (r_i m_i).
\end{align*}
The key point is how an $R$-coefficient moves across the tensor sign. Since $S$ is an $R$-module through the ring homomorphism $\varphi: R \to S$, the scalar $r_i$ acts on $s$ as multiplication by $\varphi(r_i)$. The balancing relation in $S \otimes_R M$ gives
\begin{align*}
s \otimes (r_i m_i) = (\varphi(r_i)s) \otimes m_i.
\end{align*}
Finally, the $S$-module structure on $S \otimes_R M$ is multiplication on the first tensor factor. Hence
\begin{align*}
(\varphi(r_i)s) \otimes m_i = (\varphi(r_i)s) \cdot (1_S \otimes m_i).
\end{align*}
Putting these identities together gives
\begin{align*}
s \otimes m = \sum_{i=1}^{k} (\varphi(r_i)s) \cdot (1_S \otimes m_i).
\end{align*}
This writes the arbitrary pure tensor $s \otimes m$ as an $S$-linear combination of the elements $1_S \otimes m_1,\ldots,1_S \otimes m_k$.
[/guided]
[/step]
[step:Pass from pure tensors to arbitrary elements of $S \otimes_R M$]
Let $x \in S \otimes_R M$. By the construction of the tensor product, $x$ is a finite sum of pure tensors, so there exist an integer $\ell \ge 0$, elements $s_1,\ldots,s_\ell \in S$, and elements $n_1,\ldots,n_\ell \in M$ such that
\begin{align*}
x = \sum_{j=1}^{\ell} s_j \otimes n_j.
\end{align*}
By the previous step, each pure tensor $s_j \otimes n_j$ belongs to the $S$-submodule generated by $G$. Since an $S$-submodule is closed under finite sums, $x$ belongs to the $S$-submodule generated by $G$.
Thus $G=\{1_S \otimes m_i : 1 \le i \le k\}$ generates $S \otimes_R M$ as an $S$-module. Since $G$ is finite, $S \otimes_R M$ is finitely generated as an $S$-module.
[/step]
