This is one of thousands concrete statements which are best (or in fact only) explained in the functorial approach to mathematics: Instead of looking at an object $X$ "internally", look at its represented functor $\hom(X,-)$, which is usually described by means of a universal property of $X$. Put differently: Forget about elements, think about morphisms! All this is justified by the Yoneda Lemma.
The module of differentials $\Omega^1_{S/R} \in \mathsf{Mod}(S)$ is defined by
$$\hom_S(\Omega^1_{S/R},M) = \mathrm{Der}_R(S,M)$$
for $S$-modules $M$. Now let $\phi : S \to S'$ be a surjective homomorphism "over" $\psi : R \to R'$ (diagram). Then we have for $S'$-modules $M$:
$$\hom_{S'}(\Omega^1_{S'/R'},M) = \mathrm{Der}_{R'}(S',M)$$
I am a bit lazy and don't write down the functors which restrict scalars, but they will be all over the place. Let $I := \ker(\phi)$. Then we have an isomorphism of $R$-modules $S' \cong S/I$. Hence,
$$\mathrm{Der}_{R'}(S',M) \subseteq \hom_{R'}(S',M) \subseteq \hom_{R}(S',M) \cong \{\delta \in \hom_R(S,M) : \delta|_I = 0\}.$$
The image consists of all $R$-linear maps $\delta : S \to M$ vanishing on $I$ such that the $R$-linear map $\delta' : S' \to M$ defined by $\delta' \circ \phi = \delta$ satisfies the Leibniz rule and vanishes on $R'$. This means that $\delta$ satisfies the Leibniz rule and that $\delta$ vanishes on $T := \{s \in S : \phi(s) \in R' \cdot 1\}$. This contains $I$. Thus, if $d : S \to \Omega^1_{S/R}$ is the universal differential,
$$\mathrm{Der}_{R'}(S',M) \cong \{\delta \in \mathrm{Der}_R(S,M) : \delta|_T = 0\}$$
$$\cong \{f \in \hom_S(\Omega^1_{S/R},M) : f(d(T))=0\} \cong \hom_S(\Omega^1_{S/R}/\langle d_S(T) \rangle ,M)$$
Since $\Omega^1_{S/R}/\langle d_S(T) \rangle$ is killed by $I$, it carries the structure of an $S'$-module. It follows
$$\hom_{S'}(\Omega^1_{S'/R'},-) \cong \hom_{S'}(\Omega^1_{S/R}/\langle d_S(T) \rangle,-)$$
and hence $\Omega^1_{S'/R'} \cong \Omega^1_{S/R}/\langle d_S(T) \rangle$ by Yoneda. $\square$
Try to prove this "directly" using elements in one of the explicit constructions of $\Omega^1$. Does it work at all? If yes, how many pages?
In order to illustrate the technique, let me also show a quick proof of the next lemma: If $A \to B \to C$ are homomorphisms of commutative rings, then there is an exact sequence
$$C \otimes_B \Omega^1_{B/A} \to \Omega^1_{C/A} \to \Omega^1_{C/B} \to 0.$$
Proof: By definition of a cokernel and the Yoneda Lemma, the claim is equivalent to an exact sequence
$$0 \to \hom_C(\Omega^1_{C/B},M) \to \hom_C(\Omega^1_{C/A},M) \to \hom_C(C \otimes_B \Omega^1_{B/A},M)$$
naturally in $M \in \mathsf{Mod}(C)$. By definition of $\Omega^1$ and the adjunction between scalar extension and restriction, this simplifies to
$$0 \to \mathrm{Der}_B(C,M) \to \mathrm{Der}_A(C,M) \to \mathrm{Der}_A(B,M).$$
The map on the left is the obvious inclusion, the map on the right is given by pullback with $B \to C$. Exactness is clear. $\square$
