Let $X$ be an abelian variety over an algebraically closed field, $\alpha \in \Omega_{X,0} \otimes_{\mathscr{O}_{X,0} } k(0) $ (= the dual of $T_{X,0}$) Then does this $\alpha$ extend to an element of $\Gamma(X, \Omega_X)$? And does this extension define the isomorphism $T_{X,0}^* \otimes_k \mathscr{O}_X \to \Omega_X$?
I think this is very fundamental, however I can't understand it at all. My text says: defining $\omega_\alpha (x)$ to be $t^*_{-x}(\alpha)$, we have $\omega_\alpha \in \Gamma(X, \Omega)$, the unique translation-inveriant differential form in $\Omega_{X,\xi} = \Omega_{k(X)/k}$ whose value at $0$ is $\alpha$. What is $\omega_\alpha$? How does defining $\omega_\alpha$ at each point give a section of $\Omega$?
Let $f:G\to \operatorname{Spec} k$ be a group-scheme over $\operatorname{Spec} k$ with identity $e\in G(k)$ (equivalently, $e:\operatorname{Spec} k\to G$ is the section of $f$ which picks out the identity). We show that $\Omega^1_{G/k} \cong f^*e^*\Omega_{G/k}^1$. In particular, this gives the desired isomorphism $T_{G,e}^*\otimes \mathcal{O}_{G/k} \to \Omega^1_{G/k}$.
Consider the following diagram:
$$\require{AMScd} \begin{CD} G\times G @>{pr_2}>> G\\ @VV{pr_1}V @VV{f}V \\ G @>{f}>> \operatorname{Spec} k \end{CD}$$
it is straightforward to check that the square is a pullback square. Consider now the following diagram:
$$\require{AMScd} \begin{CD} G\times G @>{m}>> G\\ @VV{pr_1}V @VV{f}V \\ G @>{f}>> \operatorname{Spec} k \end{CD}$$
Consider a morphism of the second diagram to the first which is the identity everywhere except $G\times G\to G\times G$ is given by $(pr_1,m)$. Since this is an isomorphism (with inverse $(pr_1,m\circ(i\circ pr_1,pr_2)$) and all the faces commute, we can see that in fact the second diagram is also a pullback square.
Therefore we can now compute $\Omega^1_{pr_1}$, the sheaf of relative differentials for the morphism $pr_1$, two different ways: the first square gives that $\Omega^1_{pr_1}\cong pr_2^*\Omega^1_f$, while the second square gives that $\Omega^1_{pr_1}\cong m^*\Omega^1_{f}$. So $m^*\Omega^1_{G/k}\cong pr_2^*\Omega^1_{G/k}$ as sheaves on $G\times G$. Pullback of these sheaves along $\mu = (id,e): G\to G\times G$ gives that
$$ \mu^*m^*\Omega^1_{G/k} \cong (m \circ \mu)^*\Omega^1_{G/k} \cong id^*\Omega^1_{G/k} \cong \Omega^1_{G/k}$$ and that $$ \mu^*pr_2^*\Omega^1_{G/k} \cong (pr_2 \circ \mu)^*\Omega^1_{G/k} \cong (e\circ f)^*\Omega^1_{G/k} \cong f^*e^*\Omega^1_{G/k}.$$
Thus $\Omega^1_{G/k} \cong f^*e^*\Omega^1_{G/k}$. In particular, as $e^*\Omega^1_{G/k} = T_{G,e}^*$ this defines an isomorphism $T_{G,e}^*\otimes_k \mathcal{O}_{G/k} \cong \Omega^1_{G/k}$. Additionally, by the adjunction between $f^*$ and $f_*$ and our descriptions of the maps involved in constructing the isomorphism above, the map $e^*\Omega^1_{G/k} \to f_*\Omega^1_{G/k}$ has as it's image all of the $G$-invariant 1-forms. But $e^*(f^*\alpha)=\alpha$, so $f^*\alpha$ is a $G$-invariant 1-form, and it can be seen that it satisfies the definition of $\omega_\alpha$ given in the problem statement. So $\omega_\alpha$ as defined in the problem is honestly a section of the bundle of differential forms.