I am trying to understand the proof of Lemma 130.6 in CommAlg of the stacks project. I will try to recall everything now:
Suppose we have a diagram of commutative rings
$$\require{AMScd} \begin{CD} S @>{\varphi}>> S'\\ @AA{\alpha}A @AA{\beta}A\\ R @>{\psi}>> R' \end{CD}$$
This induces a commutative diagram
$$\require{AMScd} \begin{CD} \Omega_{S/R} @>>> \Omega_{S'/R'}\\ @AA{d}A @AA{d}A\\ S @>{\varphi}>> S' \end{CD}$$
Lemma (130.6): Suppose that $\varphi : S \longrightarrow S'$ is surjective and we write $ I $ for its kernel. Then the map $ \Omega_{S/R} \longrightarrow \Omega_{S'/R'}$ is surjective and it's kernel is generated as an $S$-module by elements of the form $ ds$ where $s \in S$ such that $ \varphi(s) = \beta (r')$ for some $ r' \in R'$.
The surjectivity statement is clear to me. What I don't understand is the argument about the generators of the Kernel. In the proof the following is claimed:
Claim: A diagram chase shows that the kernel is certainly generated by elements of the form $ ids$ where $ i \in I$ and $ s \in S$ such that $ \varphi(s) = \beta (r')$ for some $ r' \in R$.
So my question is: Why is the above claim true?
Edit: Here is one idea I had that gets around the argument in the stacks project. Suppose that $ f dg \in Ker( \Omega_{S/R} \longrightarrow \Omega_{S'/R'}) $. If $ \varphi(f) = 0$ then we can write $ fdg = gdf-d(fg)$ and we win. Else if $ \varphi(f) \neq 0$ then it follows that $ \varphi(g) = \beta(r')$ for some $r' \in R'$. Is it convincing?