Let $G$ be any group scheme over $S$. Let $I\subset\mathcal{O}_G$ be the augmentation ideal defining the identity section of $G$ over $S$. Then, by definition, the first infinitesimal neighborhood of the identity in $G$ is the closed sub-scheme of $G$ defined by the ideal $I^2$. Moreover, $I/I^2$ is a coherent sheaf on $S$: it is nothing but the sheaf of invariant differentials on $G$, namely $\Omega_{G}$. In the book 'Neron Models', for example, you will find in Proposition 4.2.1 that there is a canonical isomorphism from $e^*\Omega_{G/S}$ to $\Omega_G$; here, $\Omega_{G/S}$ is the *full* sheaf of differentials on $G$ and $e:S\to G$ is the identity section.

Now, $\Omega_{G/S}$ is the ideal of the diagonal embedding of $G$ in its first order neighborhood $G^{(1)}\subset G\times_S G$. Pull everything back by the map $g\mapsto (e,g)$ from $G$ to $G\times G$. Then the diagonal embedding pulls back to the identity section and its first order neighborhood pulls back to the first order neighborhood of $e$ in $G$. So we see that $e^*\Omega_{G/S}=I/I^2$.

So $\mathcal{O}_{G}/I^2$ is non-canonically isomorphic to $\mathcal{O}_S\oplus\epsilon\Omega_G$, where $\epsilon^2=0$. This is because we have a sequence of maps
$$\mathcal{O}_S\to \mathcal{O}_G/I^2\to \mathcal{O}_G/I=\mathcal{O}_S$$
of sheaves of rings supported on the identity section. In particular, the short exact sequence
$$0\to \Omega_G\to\mathcal{O}_G/I^2\to\mathcal{O}_S\to 0$$
admits a section. Which means that we can write $\mathcal{O}_G/I^2$ as a direct sum $\mathcal{O}_{S}\oplus\Omega_G$, with the multiplication given by
$$(f,\omega)\cdot(g,\omega')=(fg,f\omega'+g\omega).$$
Writing it as $\mathcal{O}_S\oplus\epsilon\Omega_G$ is just a compressed way to indicate this.

That this pull-back is a deformation of the trivial sheaf follows from the fact that the restriction of the Poincare bundle to $G\times 0$ is canonically trivialized.

`${\cal O}_S \oplus \epsilon \Omega_{\hat G}$`

is a sheaf of rings on $S$ with $\epsilon^2 = 0$; you can locally take Spec and patch these together to produce a scheme over $S$. – Tyler Lawson Oct 20 '11 at 19:33