In chapter 9 Derivations of the book Commutative Ring Theory by H. Matsumura, there is the following situation: Given a ring $R$, and $R$-algebras $A, B, C$ with homomorphisms of $R$-algebras $f: B \rightarrow A$, $g: C \rightarrow A$, we say a morphism $h: C \rightarrow B$ is a lifting of $g$, if $f \circ h = g$.
Now let $h$ and $h'$ be two such liftings.
Let $N$ denote the kernel of $f$, i.e. we get the exact sequence $$0\rightarrow N \rightarrow B \rightarrow A.$$ If $N^2 = 0$, $N$ becomes a $f(B)$-module, and by the morphism $g: C \rightarrow f(B)$ also a $C$-module. Matsumura claims now that it is "easy to see" that $d = h-h': C \rightarrow N$ is a $R$-derivation from $C$ to $N$. I was able to show this by addding a zero at some point, but I stumbled upon another fact, and am curious if my reasoning is correct.
Claim: $d$ is $C$-module homomorphism.
Proof: Additivity is clear, and also required for a $R$-derivation.
Let $c, c' \in C$ be two elements, and put them into $d$: $$d(c\cdot c') = h(cc') - h'(cc') = h(c)h(c') - h'(c)h'(c')$$ Observe that $h(c')-h'(c')$ is itself an element in $N$, so multiplying by $h(c)$ is the same as letting $c$ act by the $C$-module structure defined above. But $h(c)$ and $h'(c)$ have the same image in $f(B) \subset A$, thus they act the same. Hence we get $$d(cc') =h(c)\cdot(h(c') - h'(c')) = c \cdot d(c')$$
The reasoning that $h(c)$ and $h'(c)$ act the same on $N$ seems valid, at least I also used it to prove that $d$ is a derivation.
If we have a derivation $D \in \text{Der}(A,M)$, where $A$ is a ring and $M$ is an $A$-module, then $D$ is a morphism of $A$-modules if and only if $D = 0$. Indeed, given $x \in A$, we have
$D(x) = D( x \cdot 1) = x \cdot D(1) = x \cdot 0 = 0.$
So I believe that there is a jump somewhere going from the expression
$h(c)h(c') - h'(c)h'(c')$
to
$h(c) \cdot \Big( h(c') - h'(c') \Big)$.
Note that $h(c)$ and $h'(c)$ only have the same action on $N$, not on $h(C)$.