If $V_n$ is $n+1$-dimensional repn of $SU(2)$ how to compute decomposition into irreducibles of $S^nV_2$ efficiently?
Doing some rather tedious computation by picking a basis in $V_2$ and considering action of diagonal representatives of classes of $SU(2)$ then computing coefficients in the character, I arrived at the result:
$$ S^nV_2 = V_{2n} \oplus V_{2n-4} \oplus V_{2n-6} \oplus \dots \oplus V_0$$
Can this be obtained using properties of the symmetric product and some clever observation? The final formula looks rather nice and it'd be nice if it had some intuitive sense...
Unfortunately, there is no nice property of the symmetric product to be used here. Your computation is the $m=2$ case of the plethysm of symmetric powers $$ S^n(S^m(V_1))\ . $$ Understanding plethysm in general is a very difficult open problem. That's for the bad news. The good news is for $SL_2$ the decomposition is known since about 1856 and it is called the Cayley-Sylvester formula. It gives the multiplicity of the irreducible module $V_k$ as $$ \left\{m,n,\frac{mn-k}{2}\right\}-\left\{m,n,\frac{mn-k}{2}-1\right\} $$ where $\{a,b,w\}$ denotes the number of integer partitions of $w$ which fit inside an $a$ by $b$ rectangle. The easiest proof is what you did, namely a character computation which essentially produces the $q$-binomials or Gaussian polynomials.
Note that in your particular case, this classically corresponds to counting covariants of degree $n$ of the generic quadratic binary form $Q$. These are linear combinations of $Q^i \Delta^j$ where $\Delta=b^2-4ac$ is the discriminant of the quadratic.