Your question is equivalent to the computation of $H^0(\mathcal{I}_S(2))$. In the example you give, $S$ is a complete intersection of $4$ quartics and so the resolution of its ideal sheaf $\mathcal{I}_S$ is given by the Koszul complex (I write $\mathcal{O}$ instead of $\mathcal{O}_{\mathbb{P}^9}$): $0 \to \mathcal{O}(-8) \to \mathcal{O}(-6)^{\oplus 4} \to \mathcal{O}(-4)^{\oplus 6} \to \mathcal{O}(-2)^{\oplus 4} \to \mathcal{I}_S \to 0$. Tensoring with $\mathcal{O}(2)$ we obtain: $0 \to \mathcal{O}(-6) \to \mathcal{O}(-4)^{\oplus 4} \to \mathcal{O}(-2)^{\oplus 6} \to \mathcal{O}^{\oplus 4} \to \mathcal{I}_S(2) \to 0$. Splitting this exact sequence into short exact ones it is immediate to check that $H^0(\mathcal{I}_S(2))=H^0(\mathcal{O}^{\oplus 4})=4$, as Algori states in his comment. Therefore the linear system of quadrics passing through $S$ has dimension $4-1=3$.