Let $\chi$ be the character of some finite-dimensional $G$-representation $V$, for $G$ a finite group. The virtual character $\chi(g^n)$ can be found using a categorified version of the cycle index of the symmetric group $S_n$:
$$\chi(g^n) = \frac{1}{n!} \sum_{\lambda} \pm C_{\lambda} \mathbb{S}_{\lambda} V,$$
where the sum runs over all partitions of $n$, $C_{\lambda}$ stands for the size of the conjugacy class associated with $\lambda$, $\mathbb{S}_{\lambda}$ is a Schur functor, and the sign depends on the parity of the number of even parts of $\lambda$. I'm sure you are familiar with the special case of this formula
$$ \chi(g^2) = \frac{1}{2} \left[ \mathbb{S}_{2} V - \mathbb{S}_{1+1} V \right] , $$
which is just another way to talk about Frobenius-Schur indicators. In fact, these characters are called "higher Frobenius-Schur indicators," which seems like a reasonable name.
Now the bad news. Nobody knows in general what Schur functors do to irreducible representations. Even in the special case of the standard representation of the symmetric group, I do not share Bruce Westbury's optimism about decomposing exterior powers into irreducibles, since every Schur functor may be written as a $\mathbb{Z}$-linear combination of tensor products of exterior power functors, and the Littlewood-Richardson coefficients would then allow us to compute arbitrary plethysm in the symmetric group.
As a last point, it is certainly false that $\chi_{\wedge^k \mbox{perm}}$ counts fixed $k$-element subsets. Anyway, it is not hard to show that the "choose k" functor from permutation representations is actually a virtual Schur functor.
For question two, you are asking about a composition of two Schur functors, i.e. a plethysm. More specifically you want to know $h_2 \circ h_k$ and $h_2 \circ e_k$. These can be found in Example 9 in the section on plethysm in Symmetric functions and Hall polynomials:
$$ h_2 \circ h_k = \sum_{j \text{ even}} s_{(2k-j,j)}$$
and
$$ h_2 \circ e_k = \sum_{j \text{ even}} s_{(k+j,k-j)^T}$$
where $(\cdot)^T$ denotes the transposed Young diagram, and the sum is taken over those even $j$ that make the subscript a valid Young diagram. These translate to universal identities between representations, i.e. an isomorphism of functors. They express how the composed functors $\mathrm{Sym}^2(\mathrm{Sym}^k(-))$ and $\mathrm{Sym}^2(\bigwedge^k(-))$ can be written as direct sums of Schur functors.
When you apply a Schur functor to the defining representation of $\mathrm{SU}(n)$, the result is nonzero if and only if the corresponding partition has at most $n$ parts. Think about how the functor $\bigwedge^k$ vanishes on any vector space of dimension less than $k$. This explains the observation you make in a comment below, that not all terms in the second sum will appear if $n$ is small.
Best Answer
Let $W = \mathbb C^n$, with the obvious representation of $S_n$.
Lemma Let $\lambda$ be a partition of $n$. The multiplicity of the irreducible representation $V_\lambda$ inside $W^{\otimes k}$ is given by the number of ways of obtaining the partition $\lambda$ by starting with the partition $[n]$, and then alternating between removing a box from the Ferrers diagram and adding a box to the Ferrers diagram $k$ times.
Proof Induction on $k$, the base case $k=0$ being trivial. We have $$ W^{\otimes (k+1)} = W^{\otimes k} \otimes \mathrm{Ind}_{S_{n-1}}^{S_n} \mathbb C = \mathrm{Ind}_{S_{n-1}}^{S_n} \mathrm{Res}^{S_n}_{S_{n-1}} W^{\otimes k},$$ see e.g. Exercise 3.16 of Fulton-Harris. Thus the result follows from Pieri's rule. QED
Let $V = W/\mathbb C$ be the irreducible representation corresponding to the partition $[n-1,1]$. Then $\wedge^k V$ is also irreducible, with corresponding partition $[n-k,1,1,\ldots,1]$, for $k\leq n-1$. By the preceding lemma, it is easy to see that the multiplicity of $\wedge^k V$ inside $W^{\otimes k}$ is exactly $1$ for $k \leq n-1$.
It follows in particular that $\wedge^k V$ occurs with multiplicity $0$ inside $\mathrm{Sym}^k(W)$ for $k=2,\ldots,n-1$. Indeed $\wedge^k(V) \subset \wedge^k(W)$ and $\wedge^k(W) \cap \mathrm{Sym}^k(W) = \{0\}$ for $k \geq 2$, where the intersection is taken inside of $W^{\otimes k}$. In particular this representation doesn't occur inside $\mathrm{Sym}^k(V)$, either.
Also, here is some SAGE code that will compute things for you.
s = SymmetricFunctions(QQ).schur() V = s([n-1,1]) ext = s([1]*k) sym = s([k]) print ext.inner_plethysm(V) ## decomposition of \wedge^k(V) print sym.inner_plethysm(V) ## decomposition of Sym^k(V)