By composition we mean the map $$ R\SheafHom_{\mathcal{O}_Y}(\mathcal{O}_X, \mathcal{O}_X) \otimes_{\mathcal{O}_Y}^\mathbf{L} R\SheafHom_{\mathcal{O}_Y}(\mathcal{O}_X, \mathcal{O}_X) \longrightarrow R\SheafHom_{\mathcal{O}_Y}(\mathcal{O}_X, \mathcal{O}_X) $$ of Cohomology, Lemma \ref{cohomology-lemma-internal-hom-composition}. This induces multiplication maps $$ \SheafExt^a_{\mathcal{O}_Y}(\mathcal{O}_X, \mathcal{O}_X) \otimes_{\mathcal{O}_Y} \SheafExt^b_{\mathcal{O}_Y}(\mathcal{O}_X, \mathcal{O}_X) \longrightarrow \SheafExt^{a + b}_{\mathcal{O}_Y}(\mathcal{O}_X, \mathcal{O}_X) $$ Please compare with More on Algebra, Equation (\ref{more-algebra-equation-simple-tor-product}). The statement of the lemma means that the induced map $$ \SheafExt^1_{\mathcal{O}_Y}(\mathcal{O}_X, \mathcal{O}_X) \otimes \ldots \otimes \SheafExt^1_{\mathcal{O}_Y}(\mathcal{O}_X, \mathcal{O}_X) \longrightarrow \SheafExt^i_{\mathcal{O}_Y}(\mathcal{O}_X, \mathcal{O}_X) $$ factors through the wedge product and then induces an isomorphism. To see this is true we may work locally on $Y$. Hence we may assume that we have global sections $f_1, \ldots, f_r$ of $\mathcal{O}_Y$ which generate $\mathcal{I}$ and which form a Koszul regular sequence. Denote $$ \mathcal{A} = \mathcal{O}_Y\langle \xi_1, \ldots, \xi_r\rangle $$ the sheaf of strictly commutative differential graded $\mathcal{O}_Y$-algebras which is a (divided power) polynomial algebra on $\xi_1, \ldots, \xi_r$ in degree $-1$ over $\mathcal{O}_Y$ with differential $\text{d}$ given by the rule $\text{d}\xi_i = f_i$. Let us denote $\mathcal{A}^\bullet$ the underlying complex of $\mathcal{O}_Y$-modules which is the Koszul complex mentioned above. Thus the canonical map $\mathcal{A}^\bullet \to \mathcal{O}_X$ is a quasi-isomorphism. We obtain quasi-isomorphisms $$ R\SheafHom_{\mathcal{O}_Y}(\mathcal{O}_X, \mathcal{O}_X) \to \SheafHom^\bullet(\mathcal{A}^\bullet, \mathcal{A}^\bullet) \to \SheafHom^\bullet(\mathcal{A}^\bullet, \mathcal{O}_X) $$ by Cohomology, Lemma \ref{cohomology-lemma-Rhom-strictly-perfect}. The differentials of the latter complex are zero, and hence $$ \SheafExt^i_{\mathcal{O}_Y}(\mathcal{O}_X, \mathcal{O}_X) \cong \SheafHom_{\mathcal{O}_Y}(\mathcal{A}^{-i}, \mathcal{O}_X) $$ For $j \in \{1, \ldots, r\}$ let $\delta_j : \mathcal{A} \to \mathcal{A}$ be the derivation of degree $1$ with $\delta_j(\xi_i) = \delta_{ij}$ (Kronecker delta). A computation shows that $\delta_j \circ \text{d} = - \text{d} \circ \delta_j$ which shows that we get a morphism of complexes. $$ \delta_j : \mathcal{A}^\bullet \to \mathcal{A}^\bullet[1]. $$ Whence $\delta_j$ defines a section of the corresponding $\SheafExt$-sheaf. Another computation shows that $\delta_1, \ldots, \delta_r$ map to a basis for $\SheafHom_{\mathcal{O}_Y}(\mathcal{A}^{-1}, \mathcal{O}_X)$ over $\mathcal{O}_X$. Since it is clear that $\delta_j \circ \delta_j = 0$ and $\delta_j \circ \delta_{j'} = - \delta_{j'} \circ \delta_j$ as endomorphisms of $\mathcal{A}$ and hence in the $\SheafExt$-sheaves we obtain the statement that our map above factors through the exterior power. To see we get the desired isomorphism the reader checks that the elements $$ \delta_{j_1} \circ \ldots \circ \delta_{j_i} $$ for $j_1 < \ldots < j_i$ map to a basis of the sheaf $\SheafHom_{\mathcal{O}_Y}(\mathcal{A}^{-i}, \mathcal{O}_X)$ over $\mathcal{O}_X$.