Take a local basis $e_1,e_2$ for $E$, and then get a local basis $E_0=e_1^{2n}, E_1=e_1^{2n-1}e_2, \dots, E_{2n}=e_2^{2n}$ for $S^{2n}E$. If we scale $e_1$ by $a_1$ and $e_2$ by $a_2$, we scale $E_j$ by $a_1^{2n-j}a_2^j$. So we scale $E_0 \wedge \dots \wedge E_{2n}$ by $a_1^p a_2^p$ where $p=0+1+2+\dots+2n=(2n+1)n$. So $\det S^{2n} E = (\det E)^{\otimes (2n+1)n}$. In $S^{2n}E \otimes \det E^{\otimes -n}$, we replace each $E_j$ by something with an extra $a_1^{-n}a_2^{-n}$. There are $2n+1$ basis elements $E_j$, so overall an extra factor of $a_1^{-n(2n+1)}a_2^{-n(2n+1)}$. So $S^{2n}E \otimes \det E^{\otimes -n}$ has local section invariant under these rescalings. It is easy to see that it is also invariant under adding multiples of $e_2$ to $e_1$, and vice versa, so that gives us invariance of our section under all linear transformations of $e_1$, $e_2$.