(Notations as in Huybrechts' book.) Consider the maps $\mathrm{Hom}(C,S(B))\rightarrow\mathrm{Hom}(C,C_0)$ and $\mathrm{Hom}(A[1],C_0)\rightarrow\mathrm{Hom}(C,C_0)$, and let $I$ and $J$ be their images. Huybrechts tells you how to define linear forms $\Xi_I:I\rightarrow k$ and $\Xi_I:J\rightarrow k$, and shows that $\Xi_I|_{I\cap J}=\Xi_J|_{I\cap J}$. The latter condition says exactly that there is a linear form on $I+J$ restricting to $\Xi_I$ and $\Xi_J$; extending this form to all of $\mathrm{Hom}(C,C_0)$ gives the desired $\Xi=(\xi,-)$.

(Notations as in Huybrechts' book.) Consider the maps $\mathrm{Hom}(C,S(B))\rightarrow\mathrm{Hom}(C,C_0)$ and $\mathrm{Hom}(A[1],C_0)\rightarrow\mathrm{Hom}(C,C_0)$, and let $I$ and $J$ be their images. Huybrechts tells you how to define linear forms $\Xi_I:I\rightarrow k$ ("condition $\mathrm{i}')$") and $\Xi_J:J\rightarrow k$ ("condition $\mathrm{ii}')$"), and shows that $\Xi_I|_{I\cap J}=\Xi_J|_{I\cap J}$. The latter condition says exactly that there is a linear form on $I+J$ restricting to $\Xi_I$ on $I$ and $\Xi_J$ on $J$; extending this form to all of $\mathrm{Hom}(C,C_0)$ gives the desired form $\Xi$ Serre dual to $\xi$.