As mentioned in my comment above, the answer is given by the Gauss equations, which provide a formula for the Riemann curvature tensor in terms of the second fundamental form of a hypersurface in Euclidean space. Here is a description that is more explicit than the one I give above:

Let's view tensor products of $E$ to be multilinear functions of $E^*$. Then the map from $h \in S^2E$ to $R \in S^2\bigwedge^2 E$ is given by $$ R(v_1, v_2, v_3, v_4) = h(v_1,v_3)h(v_2,v_4) - h(v_1, v_4)h(v_2,v_3), $$ for any $v_1, v_2, v_3, v_4 \in E^*$. Note that as a bonus, $R$ also satisfies the first Bianchi identities $$ R(v_1, v_2, v_3, v_4) + R(v_2, v_3, v_1, v_4 + R(v_3, v_1, v_2, v_4) = 0. $$

As mentioned in my comment above, the answer is given by the Gauss equations, which provide a formula for the Riemann curvature tensor in terms of the second fundamental form of a hypersurface in Euclidean space. Here is a description that is more explicit than the ones I gave in my comments above:

Let's view tensor products of $E$ to be multilinear functions of $E^*$. Then the map from $h \in S^2E$ to $R \in S^2\bigwedge^2 E$ is given by $$ R(v_1, v_2, v_3, v_4) = h(v_1,v_3)h(v_2,v_4) - h(v_1, v_4)h(v_2,v_3), $$ for any $v_1, v_2, v_3, v_4 \in E^*$. Note that, as a bonus, $R$ also satisfies the first Bianchi identities $$ R(v_1, v_2, v_3, v_4) + R(v_2, v_3, v_1, v_4) + R(v_3, v_1, v_2, v_4) = 0. $$