What is the universal property of the thickening $Y[\varepsilon]$? Given an $S$-scheme $Y$, let $Y[\varepsilon]$ denote thickening $Y[\varepsilon]=Y \times_S D_S$ of $Y$. Here $D_S$ is the $S$-scheme $D\times_\mathbb Z S\to S$ where $D = \operatorname{Spec} \mathbb{Z}[\varepsilon]/(\varepsilon^2)$. The $S$-scheme $Y[\varepsilon]$ should have a mapping out universal property in $\mathrm{Sch}_{/S}$, which is probably related to derivations. Unfortunately I cannot find a good reference. A morphism $f \in \mathrm{Hom}_S(Y[\varepsilon],X)$ is the same thing as a map $f \in \mathrm{Hom}_S(Y,X)$ plus what data? Help is appreciated.
Context: I am trying to understand the proof of Lemma 5.12.1 in Martin Brandenburg’s PhD thesis.
 A: I only know how to say things about affine schemes so please permit me to restrict my attention to that case. Let $k$ be a commutative ring and $f : R \to S$ be a morphism of commutative $k$-algebras. An $f$-derivation (I'm making this up, I don't know what the standard term is here) is a $k$-linear map $D : R \to S$ satisfying the Leibniz rule
$$D(ab) = f(a) D(b) + D(a) f(b).$$
You can check that the data of a pair $(f, D)$ consisting of a morphism $f$ and an $f$-derivation $D$ is the same thing as the data of a morphism $R \to S[\varepsilon]/\varepsilon^2$; the morphism associated to such a pair is
$$R \ni a \mapsto f(a) + D(a) \varepsilon \in S$$
and the Leibniz rule is exactly equivalent to the condition that this map respects multiplication. Also equivalently, this is the data of a $k[\varepsilon]/\varepsilon^2$-linear morphism $R[\varepsilon]/\varepsilon^2 \to S[\varepsilon]/\varepsilon^2$.
Intuitively you can think of an $f$-derivation as an element of the tangent space of $f^{\ast} : \text{Spec } S \to \text{Spec } R$ in the "Hom scheme" of morphisms $\text{Spec } S \to \text{Spec } R$, and thickening is just a globalization of this construction.
As a special case, you can check that a morphism $f : R \to k[\varepsilon]/\varepsilon^2$ is the same thing as a pair of a $k$-point of $\text{Spec } R$ and a Zariski tangent vector at this $k$-point (at least if $k$ is algebraically closed). This means $\text{Spec } k[\varepsilon]/\varepsilon^2$ is the "walking tangent vector," so if you want to compute a tangent vector to a morphism $f : X \to Y$ of $k$-schemes, that means you want to compute
$$\text{Hom}(\text{Spec } k[\varepsilon]/\varepsilon^2, \text{Hom}(X, Y)) \cong \text{Hom}(\text{Spec } k[\varepsilon]/\varepsilon^2 \times X, Y)$$
where the "Hom scheme" $\text{Hom}(X, Y)$ may not always exist as a scheme but always exists at least as a presheaf $\text{Hom}(- \times X, Y)$.
Edit: This also means that the "Hom scheme" $\text{Hom}(\text{Spec } k[\varepsilon]/\varepsilon^2, X)$ itself can be interpreted as the tangent bundle of $X$ which is roughly the conceptual content of Lemma 5.12.1.
A: Thank you Qiaochu Yuan for your answer! I will, for completeness, glue your answer to the non-affine case. The key insight has been that a morphism $(\phi^\#, D) \colon \mathcal{O}_X \to \phi_\ast \mathcal{O}_Y[\varepsilon]/(\varepsilon^2) = \phi_\ast \mathcal{O}_Y \oplus \phi_\ast \mathcal{O}_Y \varepsilon$ is a morphism of $p^{-1} \mathcal{O}_S$-algebras if and only if $D$ is a $p^{-1} \mathcal{O}_S$-derivation. Knowing this, the computation looks like this:
\begin{align}
  \mathrm{Hom}_S(Y, T(X/S))
  &= \mathrm{Hom}_S(Y[\varepsilon], X) \\[0.5em]
  &= \coprod_{\phi \in \mathrm{Hom}_S(Y, X)} \mathrm{Hom}_{\mathrm{Alg}(p^{-1}\mathcal{O}_S)}(\mathcal{O}_X, (\phi_\ast \mathcal{O}_Y)[\varepsilon]/(\varepsilon^2)) \\[0.5em]
  &= \coprod_{\phi \in \mathrm{Hom}_S(Y, X)} \mathrm{Der}_{p^{-1} \mathcal{O}_S}(\mathcal{O}_X, \phi_\ast \mathcal{O}_Y) \\[0.5em]
  &= \coprod_{\phi \in \mathrm{Hom}_S(Y, X)} \mathrm{Hom}_{\mathcal{O}_X}(\Omega_{X/S}, \phi_\ast \mathcal{O}_Y) \\[0.5em]
  &= \coprod_{\phi \in \mathrm{Hom}_S(Y, X)} \mathrm{Hom}_{\mathrm{Alg}(\mathcal{O}_X)}(\operatorname{Sym} \Omega_{X/S}, \phi_\ast\mathcal{O}_Y) \\[0.5em]
  &= \mathrm{Hom}_S(Y, \operatorname{Spec}_X \operatorname{Sym} \Omega_{X/S}) \,.
\end{align}
