How Appell-Humbert theorem works in the simplest case of an elliptic curve Line bundles on complex tori $V/\Lambda$ could be described by a pair $(H, \chi)$, where $H$ is a hermitian form on $V$ s.t. $\operatorname{Im} H(\Lambda, \Lambda) \subset \mathbb{Z}$, and $\chi$ is a semicharacter i.e. satisfy condition
$$
\chi(\lambda + \mu) = \chi(\lambda) \chi(\mu) exp( i \pi \operatorname{Im} H(\lambda, \mu)),
$$
for any $\lambda, \mu \in \Lambda$.
Let denote $L(H, \chi)$ the corresponding line bundle of Appell-Humbert theorem. 
For simplest case of an elliptic curve $E = V/\Lambda$ and line bundle $L=\mathcal{O}_E(p)$, ($p \in E$ is a divisor consisting of just one point) how to find explicitly pair $(H, \chi)$ for this $L$?

Well, first you need a Hermitian form $H:\mathbb{C}\times\mathbb{C}\to\mathbb{C}$ such that $\Im H(\Lambda\times\Lambda)\subseteq\mathbb{Z}$. Using a little bit of linear algebra, it is easy to see that there is only one $\mbox{mod }\mathbb{Z}$. Explicitly, if $\Lambda=\langle 1,\tau\rangle$ for $\tau\in\mathbb{H}$, then every Hermitian form that satisfies what I just wrote is an integer multiple of
$$H_0:(z,w)\mapsto \frac{z\overline{w}}{\Im\tau}.$$
We see that the map $\chi_0(a+b\tau)=(-1)^{ab\Im H_0(1,\tau)}$ is a semi-character for $H_0$, and actually all of them are of the form
$$\chi_x:a+b\tau\mapsto\chi_0e^{2\pi i\Im H_0(x,\cdot)}$$
for a certain $x\in\mathbb{C}$ (actually we can think of $x\in\mathbb{C}/\Lambda$). In your case, we can write
$$\mathcal{O}_E(p)\simeq L(H_0,\chi_p).$$
Take a look at Birkenhake-Lange, page 48, for example.
Edit: We can see this all explicitly as part of the exponential sequence:
$$0\to E\stackrel{i}{\to}\mbox{Pic}(E)\stackrel{c_1}{\to}\mathbb{Z}\to0$$
where $i(x)=L(0,e^{2\pi i\Im H_0(x,\cdot)})$ and $c_1(L(mH_0,\chi))=m$.
