Over any field, $T_0 J$ is canonically isomorphic to $H^1(C, \mathcal{O})$ and hence, by Serre duality, is dual to $H^0(C, \Omega^1)$. Seeing that $T_0 J \cong H^0(C, \mathcal{O})$ is a good example of using the functorial description of $J$.
Let $D = \mathrm{Spec}\ k[\epsilon]/\epsilon^2$. Then $T_0 J$ is the set of maps $D$ to $J$ such that the composite $\mathrm{Spec}\ k \to D \to J$ lands on $0$. For any $k$-scheme $S$, $\mathrm{Hom}(S, J)$ is the group of isomorphism classes of line bundles on $C \times S$, modulo line bundles pulled back from $S$. In the case of $D$, all line bundles on $D$ are trivial, so $\mathrm{Hom}(D, J)$ is isomorphism classes of line bundles on $D \times C$. The homs which send $\mathrm{Spec} \ k$ to $0$ are those where the line bundle on $(\mathrm{Spec} \ k) \times C$ is trivial.
Let $U = \mathrm{Spec} \ A$ be an open affine of $C$, so $D \times U$ is an open affine of $D \times C$. Note that $D \times U \cong \mathrm{Spec} A[\epsilon]/\epsilon^2$. Then we have a short exact sequence $$0 \to A_{+} \to (A[\epsilon]/\epsilon^2)^{\ast} \to A^{\ast} \to 0.$$ Here $B^{\ast}$ is the unit group of the ring $B$, and $A_{+}$ is $A$ considered as an additive group. The first map is $a \mapsto 1+a \epsilon$. This turns into a short exact sequence of sheaves on $C$: $$0 \to \mathcal{O}_C \to \mathcal{O}_{D \times C}^{\ast} \to \mathcal{O}_C^{\ast} \to 0.$$ (Note that $C$ and $D \times C$ are the same underlying topological space, so all of these are sheaves on the same space.)
So we have $H^1(C, \mathcal{O}) \to \mathrm{Pic}(D \times C) \to \mathrm{Pic}(C)$. We argued before that $T_0(J)$ is the kernel of the second map, and a little more work shows that the first map is injective, so $T_0(J) \cong H^1(C, \mathcal{O})$ as desired. "$\square$"
The end of proof symbol is in quotes, because one should really check that this isomorphism is compatible with the vector space structures and work out how it relates to endomorphisms of $C$, but this is the idea.