We first write $(R \to S, M)$ as the directed colimit of a system $(R_\lambda \to S_\lambda, M_\lambda)$ as in as in Lemma \ref{lemma-limit-module-finite-presentation}. Let $\mathfrak q \subset S$ be a prime. Let $\mathfrak p \subset R$, $\mathfrak q_\lambda \subset S_\lambda$, and $\mathfrak p_\lambda \subset R_\lambda$ the corresponding primes. As seen in the proof of Theorem \ref{theorem-openness-flatness} $$ ((R_\lambda)_{\mathfrak p_\lambda}, (S_\lambda)_{\mathfrak q_\lambda}, (M_\lambda)_{\mathfrak q_{\lambda}}) $$ is a system as in Lemma \ref{lemma-limit-module-essentially-finite-presentation}, and hence by Lemma \ref{lemma-colimit-eventually-flat} we see that for some $\lambda_{\mathfrak q} \in \Lambda$ for all $\lambda \geq \lambda_{\mathfrak q}$ the module $M_\lambda$ is flat over $R_\lambda$ at the prime $\mathfrak q_{\lambda}$. \medskip\noindent By Theorem \ref{theorem-openness-flatness} we get an open subset $U_\lambda \subset \Spec(S_\lambda)$ such that $M_\lambda$ flat over $R_\lambda$ at all the primes of $U_\lambda$. Denote $V_\lambda \subset \Spec(S)$ the inverse image of $U_\lambda$ under the map $\Spec(S) \to \Spec(S_\lambda)$. The argument above shows that for every $\mathfrak q \in \Spec(S)$ there exists a $\lambda_{\mathfrak q}$ such that $\mathfrak q \in V_\lambda$ for all $\lambda \geq \lambda_{\mathfrak q}$. Since $\Spec(S)$ is quasi-compact we see this implies there exists a single $\lambda_0 \in \Lambda$ such that $V_{\lambda_0} = \Spec(S)$. \medskip\noindent The complement $\Spec(S_{\lambda_0}) \setminus U_{\lambda_0}$ is $V(I)$ for some ideal $I \subset S_{\lambda_0}$. As $V_{\lambda_0} = \Spec(S)$ we see that $IS = S$. Choose $f_1, \ldots, f_r \in I$ and $s_1, \ldots, s_n \in S$ such that $\sum f_i s_i = 1$. Since $\colim S_\lambda = S$, after increasing $\lambda_0$ we may assume there exist $s_{i, \lambda_0} \in S_{\lambda_0}$ such that $\sum f_i s_{i, \lambda_0} = 1$. Hence for this $\lambda_0$ we have $U_{\lambda_0} = \Spec(S_{\lambda_0})$. This proves (1). \medskip\noindent Proof of (2). Let $(R_0 \to S_0, M_0)$ be as in (1) and suppose that $R = \colim R_\lambda$. Since $R_0$ is a finite type $\mathbf{Z}$ algebra, there exists a $\lambda$ and a map $R_0 \to R_\lambda$ such that $R_0 \to R_\lambda \to R$ is the given map $R_0 \to R$ (see Lemma \ref{lemma-characterize-finite-presentation}). Then, part (2) follows by taking $S_\lambda = R_\lambda \otimes_{R_0} S_0$ and $M_\lambda = S_\lambda \otimes_{S_0} M_0$. \medskip\noindent Finally, we come to the proof of (3). Let $(R_\lambda \to S_\lambda, M_\lambda)$ be as in (3). Choose $(R_0 \to S_0, M_0)$ and $R_0 \to R$ as in (1). As in the proof of (2), there exists a $\lambda_0$ and a ring map $R_0 \to R_{\lambda_0}$ such that $R_0 \to R_{\lambda_0} \to R$ is the given map $R_0 \to R$. Since $S_0$ is of finite presentation over $R_0$ and since $S = \colim S_\lambda$ we see that for some $\lambda_1 \geq \lambda_0$ we get an $R_0$-algebra map $S_0 \to S_{\lambda_1}$ such that the composition $S_0 \to S_{\lambda_1} \to S$ is the given map $S_0 \to S$ (see Lemma \ref{lemma-characterize-finite-presentation}). For all $\lambda \geq \lambda_1$ this gives maps $$ \Psi_{\lambda} : R_\lambda \otimes_{R_0} S_0 \longrightarrow R_\lambda \otimes_{R_{\lambda_1}} S_{\lambda_1} \cong S_\lambda $$ the last isomorphism by assumption. By construction $\colim_\lambda \Psi_\lambda$ is an isomorphism. Hence $\Psi_\lambda$ is an isomorphism for all $\lambda$ large enough by Lemma \ref{lemma-colimit-category-fp-algebras}. In the same vein, there exists a $\lambda_2 \geq \lambda_1$ and an $S_0$-module map $M_0 \to M_{\lambda_2}$ such that $M_0 \to M_{\lambda_2} \to M$ is the given map $M_0 \to M$ (see Lemma \ref{lemma-module-map-property-in-colimit}). For $\lambda \geq \lambda_2$ there is an induced map $$ S_\lambda \otimes_{S_0} M_0 \longrightarrow S_\lambda \otimes_{S_{\lambda_2}} M_{\lambda_2} \cong M_\lambda $$ and for $\lambda$ large enough this map is an isomorphism by Lemma \ref{lemma-colimit-category-fp-modules}. This implies (3) because $M_0$ is flat over $R_0$.