In an Algebraic Geometry course, I've seen the definition of the $A$-module of Kähler differentials $\Omega_A^1$ given a $k$-algebra $A$ ($k$ is a field). Then if $X$ is a variety, we defined the sheaf of 1-forms $\Omega_X^1$ on $X$ as the sheafification of the presheaf $\Omega^{1,p}_X$ that maps an open set $U \subseteq X$ to the $\mathcal{O}_X(X)$-module $\Omega^1_{\mathcal{O}_X(U)}$ and that maps an inclusion $V \subseteq U$ to the map $\Omega^1_{\mathcal{O}_X(U)} \to \Omega^1_{\mathcal{O}_X(V)}$ induced by the $k$-algebra map that is restriction from $\mathcal{O}_X(U)$ to $\mathcal{O}_X(V)$. This prehseaf seems to me like a composition of the two functors $\mathcal{O}_X$ and $\Omega^1$, but in order to show this I need to precisely define $\Omega^1$ as a functor.
We have that $\Omega^1$ takes a $k$-algebra $A$ and sends it to an $A$-module $\Omega^1_A$. Since the ring of scalars over which this is a module depends on the $k$-algebra itself, we can't define $\Omega^1$ as a functor $k\text{-alg} \to \text{R-Mod}$ for a fixed ring $R$. Instead it seems it goes into a general category of modules, without fixing the ring of scalars. (Update from the future: I later realised that we can just say that the codomain is the category of $k$-modules, i.e. $k$-vector spaces, but I'm still interested in the question below.)
Can we see $\Omega^1$ as a functor from $k\text{-alg} \to \text{Mod}$ where $\text{Mod}$ is the category described below?
I made my own candidate definition of $\text{Mod}$ which turned out to be the same as the one mentioned here.
Define $\mathrm{Mod}$ to be a category whose objects are modules over any ring, i.e. tuples of the form $(M, A, \cdot)$ where $M$ is an abelian group (the operation is implicit), $A$ is a ring, and $\cdot \colon A \times M \to M $ is a scalar multiplication satisfying the axioms of a module. The morphisms of $\mathrm{Mod}$ from $(M, A, \cdot)$ to $(N, B, \star)$ consist of pairs $(\phi, \sigma)$ where $\sigma\colon A \to B$ is a ring homomorphism and $\phi \colon M \to N$ is an $A$-module homomorphism if we view $N$ as an $A$-module $(N, A, \cdot_N)$ by setting $a \cdot_N n := \sigma(a) \star n$ for all $a \in A$ and $n \in N$. Composition of morphisms are componentwise compositions on such pairs.
Yes. To be a bit more precise, let $k$ be any commutative ring, let $\mathbf{CAlg}_k$ denote the category of commutative $k$-algebras, and let $\mathbf{ModCAlg}_k$ (no official notation) be the category of pairs $(A,M)$ where $A$ is a commutative $k$-algebra and $M$ is a left $A$-module. A morphism $(A,M) \to (B,N)$ consists of an algebra map $\sigma : A \to B$ and an $A$-module map $M \to \sigma^* N$ (where $\sigma^*$ denotes restriction of scalars); equivalently a $B$-module map $B \otimes_A M \to N$.
There is a forgetful functor $$P : \mathbf{ModCAlg}_k \longrightarrow \mathbf{CAlg}_k.$$ Then Kähler differentials provide a functor (let me write $\Omega^1(A)$ instead of $\Omega^1_A$) $$\widetilde{\Omega^1}: \mathbf{CAlg}_k \to \mathbf{ModCAlg}_k,\quad A \mapsto (A,\Omega^1(A))$$ such that $$P \circ \widetilde{\Omega^1} = \mathrm{id}_{\mathbf{CAlg}_k}.$$ If $\sigma : A \to B$ is a morphism of $k$-algebras, we get an $A$-linear map $$\Omega^1(\sigma) : \Omega^1(A) \to \sigma^* \Omega^1(B)$$ corresponding, via the universal property of $\Omega^1(A)$, to the $A$-derivation $$A \xrightarrow{\sigma} B \xrightarrow{d} \Omega^1(B).$$ That is, $\Omega^1(\sigma)(d(a))=d(\sigma(a))$.
By the way, you can define $\Omega^1_{X/S}$ for every morphism of ringed spaces $X \to S$.