= Solution
For a <group homomorphism> $h:A\to B$, the dual morphism $\widehat h:\widehat B\to\widehat A$ is pullback of algebraically trivial <line bundles>. Translations satisfy $h\circ T_x=T_{h(x)}\circ h$, and therefore
$$
\phi_{h^*L}(x)=h^*(T_{h(x)}^*L\otimes L^{-1})=(\widehat h\circ\phi_L\circ h)(x).
$$
Also $\phi_{L\otimes M}=\phi_L+\phi_M$ and $\phi_{L^{-1}}=-\phi_L$. Dualizing is additive: $\widehat{f+g}=\widehat f+\widehat g$, since algebraically trivial <line bundles> obey the <Theorem of the square>. Apply these identities to the three factors defining $D_L$. The two diagonal terms cancel in the expansion
$$
\begin{aligned}
\phi_{D_L(f,g)}&=(\widehat f+\widehat g)\phi_L(f+g)-\widehat f\phi_Lf-\widehat g\phi_Lg\\
&=\widehat f\phi_Lg+\widehat g\phi_Lf.
\end{aligned}
$$
\b[Hence $\boxed{\phi_{D_L(f,g)}=\widehat f\circ\phi_L\circ g+\widehat g\circ\phi_L\circ f}$.]
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