Projections to the second summands define a morphism from that complex to the complex $$ F \stackrel{s}\to F(D)\tag{*} $$ of locally free sheaves. The cone of this morphism is the complex $$ 0 \to E \stackrel{s}\to E(D) \stackrel{\rho_E}\to E(D)\vert_D \to 0 $$ which is acyclic. Therefore, $(*)$ is a locally free resolution of the original complex.
EDIT. Alternatively, let $$ K = \mathrm{Ker}\Big((\rho_E, \rho_F) \colon E(D) \oplus F(D) \to E(D)\vert_D\Big). $$ Then $K$ is locally free, the morphism $(s, s) \colon E \oplus F \to E(D) \oplus F(D)$ factors through $K$, and the complex $$ E \oplus F \to K $$ is quasiisomorphic to the original one (via the morphism, which is identical on $E \oplus F$ and is the natural embedding on $K$).