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 $4-1=3$.

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 $4$ quadrics 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 $4-1=3$.