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By your definition there is an exact sequence $$ 0 \to M^\vee \to V^\vee \to V_1^\vee \otimes \mathcal{O}/x \oplus V_3^\vee \otimes \mathcal{O}/y \to 0 $$ on $X \times \mathbb{A}^2$ (where $V$ and $V_i$ denote the pullbacks of the same named bundles). Restricting to $X \times \{x = 0\}$ one obtains an exact sequence $$ 0 \to V_1^\vee \otimes \mathcal{O}/x \to M^\vee \otimes \mathcal{O}/x \to V^\vee \otimes \mathcal{O}/x \to V_1^\vee \otimes \mathcal{O}/x \oplus V_3^\vee \otimes \mathcal{O}/(x,y) \to 0. $$ The morphism $V^\vee \otimes \mathcal{O}/x \to V_1^\vee \otimes \mathcal{O}/x$ is the restriction of the morphism dual to $V_1 \to V$, therefore it is surjective and its kernel is $V_2^\vee \otimes \mathcal{O}/x$. This means that $M^\vee\vert_{x = 0}$ is an extension of the kernel of the morphism $$ V_2^\vee\vert_{x = 0} \to V_3^\vee\vert_{x = y = 0} $$ (the morphism is induced by the composition $V_3 \to V \to V_2$) by the bundle $V_1^\vee\vert_{x = 0}$.