Derivations are maps out of Ω

Bijectivity. By [F1] the pair $({\mathrm{\Omega }}_{B\mathrm{/}A},\mathrm{d})$ is a Kähler differential module for $A\to B$, so [F2] gives, for every $B$-module $M$, that $g\mapsto g\circ \mathrm{d}$ is a bijection ${\mathrm{Hom}}_B({\mathrm{\Omega }}_{B\mathrm{/}A},M)\to {\mathrm{Der}}_A(B,M)$; the same statement holds for any Kähler differential module, since any two are related by a unique compatible isomorphism identifying the two assignments.

Additivity and $B$-linearity of the bijection. Both sides are $B$-modules: ${\mathrm{Hom}}_B({\mathrm{\Omega }}_{B\mathrm{/}A},M)$ under pointwise operations, and ${\mathrm{Der}}_A(B,M)$ under the operations of [F3]. For $g,g^{\mathrm{\prime }}\in {\mathrm{Hom}}_B({\mathrm{\Omega }}_{B\mathrm{/}A},M)$ and $c\in B$ one has $(g+g^{\mathrm{\prime }})\circ \mathrm{d}=g\circ \mathrm{d}+g^{\mathrm{\prime }}\circ \mathrm{d}$ and $(c\cdot g)\circ \mathrm{d}=c\cdot (g\circ \mathrm{d})$ as maps $B\to M$, because evaluation at any $b$ gives $cg(\mathrm{d}b)$ on both sides. Hence $g\mapsto g\circ \mathrm{d}$ is a homomorphism of $B$-modules.

Naturality. Let $t\text{\hspace{0.17em}⁣}:M\to N$ be $B$-linear. By [F3] the composite $t\circ D$ is a derivation for every $D\in {\mathrm{Der}}_A(B,M)$ and the assignment $D\mapsto t\circ D$ is $B$-linear; moreover $t\circ (g\circ \mathrm{d})=(t\circ g)\circ \mathrm{d}$ for every $B$-linear $g\text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{B\mathrm{/}A}\to M$, since both sides send $b$ to $t(g(\mathrm{d}b))$. Thus applying $t$ after the isomorphism agrees with applying $t$ before it, and the isomorphism of step 1.2 is natural in $M$: the $B$-module ${\mathrm{\Omega }}_{B\mathrm{/}A}$ represents the functor $M\mapsto {\mathrm{Der}}_A(B,M)$ by [F2].