Multiplication in Deligne cohomology: explicit formula for $p=q=1$

There is indeed a "universal" holomorphic bundle with connection on $\mathbb{C}^\times \times \mathbb{C}^\times$ which induces the bundles $r(f,g)$ defined in Esnault-Viehweg. This universal bundle has been constructed by D. Ramakrishnan using the Heisenberg group (Bulletin AMS vol. 5 n. 2, 1981).

It is nicely explained in R. Hain, Classical polylogarithms. Put \begin{equation*} H_{\mathbb{C}} = \begin{pmatrix} 1 & \mathbb{C} & \mathbb{C} \\ 0 & 1 & \mathbb{C} \\ 0 & 0 & 1 \end{pmatrix} \end{equation*} and \begin{equation*} H_{\mathbb{Z}} = \begin{pmatrix} 1 & \mathbb{Z}(1) & \mathbb{Z}(2) \\ 0 & 1 & \mathbb{Z}(1) \\ 0 & 0 & 1 \end{pmatrix}. \end{equation*} The exponential map gives a canonical projection $H_{\mathbb{Z}}\backslash H_{\mathbb{C}} \to \mathbb{C}^\times \times \mathbb{C}^\times$ with fiber $\mathbb{C}/\mathbb{Z}(2) \cong \mathbb{C}^\times$, which is the bundle you want. This bundle is not trivial but becomes trivial after pulling-back to $\mathbb{C} \times \mathbb{C}$ using this exponential map.

Denoting by $u,v$ the coordinates on $\mathbb{C} \times \mathbb{C}$ (they are just the matrix coefficients using the Heisenberg description), the pull-back of the connection is given by \begin{equation*} \nabla s = ds - s \cdot u dv/2\pi i \end{equation*} This defines a connection on $\mathbb{Z}(2)\backslash H_{\mathbb{C}}$ which descends to $H_{\mathbb{Z}} \backslash H_{\mathbb{C}}$.

Not a complete answer, but addressing the second point.

Since $(\mathbb{C}^\times)^2$ is Stein it means that topological and holomorphic classification of line bundles coincides, not that there are no nontrivial ones. $(\mathbb{C}^\times)^2$ is homotopy equivalent to the real 2d torus $(S^1)^2$, which has the volume form $d\theta_1\wedge d\theta_2$, which is the curvature (up to proportionality constant) of a connection on the line bundle corresponding to the generator of $H^2((S^1)^2,\mathbb{Z}) \simeq \mathbb{Z}$. We can take this connection to be $\theta_1 \mathrm{d}\theta_2$.

Since $H^2((S^1)^2,\mathbb{Z}) \simeq H^2((\mathbb{C}^\times)^2,\mathbb{Z}) \simeq H^1((\mathbb{C}^\times)^2,\mathcal{O}^\times)$, we get a unique holomorphic line bundle on $(\mathbb{C}^\times)^2$ whose underlying topological line bundle is the one pulled back along the retraction $(\mathbb{C}^\times)^2 \to (S^1)^2$.