This question is a follow up to Why is the norm map dual to restriction under Tate local duality?

Let $A$ and $B$ be dual abelian schemes over a base scheme $S$. For an integer $n \ge 1$, consider the Cartier duality cup product pairing $$ H^1(S, B[n]) \times H^1(S, A[n]) \rightarrow H^2(S, \mathbb{G}_m) $$ (which agrees with the edge-Ext-pairing by interpreting $B[n]$ as $\mathscr{Hom}(A[n], \mathbb{G}_m)$ and using the edge map $H^1(S, B[n]) \rightarrow \mathrm{Ext}^1(A[n], \mathbb{G}_m)$ --- see 1970-71 papers of Gamst & Hoechsmann for this agreement). Consider also the "Tate duality pairing" $$ B(S) \times H^1(S, A) \rightarrow H^2(S, \mathbb{G}_m) $$ defined by using the identification $B(S) = \mathrm{Ext}^1(A, \mathbb{G}_m)$ and the Ext-product again. The two pairings are related by the connecting homomorphism $$ B(S) \rightarrow H^1(S, B[n]) $$ and by $$ H^1(S, A[n]) \rightarrow H^1(S, A). $$ Why are the two pairings compatible? I.e., why does the obvious diagram commute? If possible, it would also be interesting to discuss the relation of the second pairing with the biextension cup product pairing $$ H^0(S, B) \times H^1(S, A) \rightarrow H^1(S, B \otimes^{\mathbb{L}} A) \rightarrow H^2(S, \mathbb{G}_m). $$ If it makes a difference, feel free to assume that $S$ is the spectrum of a nonarchimedean local field.

The compatibility that I am asking about is used in III.7.8 of Milne's "Arithmetic Duality Theorems" in the proof of Tate local duality for abelian varieties. It is also taken for granted whenever needed in a number of other places in the literature.