Functor representing the tangent space of $X \to S$ (Algebraic Geometry)

Let $X \to S$ be a morphism of schemes and define the following functor $$F: (\operatorname{Aff}/X)^{\operatorname{opp}}\to \operatorname{Set}$$ $$(f:\operatorname{Spec}A \to X) \mapsto \{ u:\operatorname{Spec}A[\epsilon] \to X | u \circ \pi=f\}$$ where $\pi$ is the associated map to the canonical projection $A[\epsilon] \to A$. I want to show that this functor is representable by the relative spectra of the sheaf of $\mathcal{O}_X$-algebras $\operatorname{Sym}\Omega^1_{X/S}$. This is pretty straightforward if we assume $X,S$ to be affine.

I would like to find a way to reduce the problem to this simpler case. To do this I fix a covering $\{U_i\}_{i \in I}$ of $X$ by affine opens such that the image of every $U_i$ lies in some affine open of $S$. Then I define the following family of open subfunctors of $F$ via, $$F_i= F \times_{h_X} h_{U_i}$$ where $h_X= \operatorname{Hom}_{ \operatorname{Sch/U}}(-,U)$. Let us remark that $h_X$ is terminal in the category of schemes over $X$ and that we have a map induced by the inclusion of $U_i$ into $X$ namely $h_{U_i} \to h_X$. Given $f:\operatorname{Spec}A \to X$ then we have that $$F_i(f:\operatorname{Spec}A \to X)=\emptyset \text{ if f does not factor through } U_i$$ and $$F_i(f:\operatorname{Spec}A \to X)=F(f:\operatorname{Spec}A \to X) \text{ otherwise.}$$ If $f:\operatorname{Spec}A \to X$ factors through $U_i$ then if I show that every $u \in F(f:\operatorname{Spec}A \to X)$ must factor through $U_i$ then I can reduce it to the affine case. However I don't see how I could show this factorization despite the fact that the topological spaces of $\operatorname{Spec}A ,\operatorname{Spec}A[\epsilon]$ are homeomorphic.

Can one show this? Is there any other way of producing a good reduction to the affine case?