There are several possible interpretations of $\Omega^k$ (admittedly some don't quite align with what you ask): ordered/unordered subsets of $k$ distinct/not necessarily distinct elements of $[n]$. There are two questions about them, "what is the character?" and "what are the multiplicities of irreducibles?".
Case 1: ordered, not necessarily distinct
In this case we can recognise the permutation representation as just the $k$-th tensor power of $\mathbb{C}^n$ (with the usual action). Since the character of $\mathbb{C}^n$ on an element of cycle type $\mu$ is just $m_1(\mu)$ (the number of parts of size $1$ in $\mu$), the character of $\Omega^k$ is $m_1(\mu)^k$. The multiplicity rules can be easily deduced from the rule for taking the tensor product with $\mathbb{C}^n$. Details can be found in RSK Insertion for Set Partitions and Diagram Algebras and the answer is "vacillating tableaux of shape $\lambda$ and size $k$". (Caveat: you need to be a little more careful if $n$ is smaller than $2k$, for example if $n=1$, then any tensor power of $\mathbb{C}^n$ is still just $\mathbb{C}^n$, which is definitely not true if $n > 1$. Feel free to ask for more detail if Secion 2 of the paper doesn't address your concerns.)
Case 2: unordered, not necessarily distinct
In this case, we can recognise $\Omega^k$ as the $k$-th symmetric power of $\mathbb{C}^n$. One way to view this is as the restriction of $\mathrm{Sym}^k(\mathbb{C}^n)$ from $GL_n(\mathbb{C})$ to $S_n$ (viewed as the subgroup of permutation matrices). Hence the character is obtained by evaluating the Schur polynomial $s_k(x_1, x_2, \ldots, x_n)$ at the eigenvalues of a permutation matrix (each $r$-cycle contributes the set of all $r$-th roots of unity). You can read more about this approach in Symmetric group characters as symmetric functions. As for the multiplicities, there is a (well-known?) formula which can be found in Exercise 7.74 of Enumerative Combinatorics Vol. 2 which states that the multiplicity of the irreducible $S^\mu$ in the $GL_n(\mathbb{C})$ irreducible indexed by $\lambda$ is
$$\langle s_\lambda, s_\mu[1 + h_1 + h_2 + \cdots ] \rangle$$
where $s_\lambda, s_\mu$ are Schur functions, square brackets denote plethysm, and $h_i$ are complete symmetric functions. In our case, $\lambda = (k)$ and we can use some tricks (which I can elaborate on, if requested) to deduce that the multiplicity is the number of semi-standard Young tableau of shape $\mu$ and weight $\nu$, such that $0 \nu_1 + 1 \nu_2 + 2 \nu_3 + \cdots = k$.
Case 3: ordered, distinct
Note first of all that if we require distinctness, $k \leq n$. As Mark Wildon pointed out in the comments, we may recognise $\Omega^k$ as the permutation module $M^{(n-k, 1^k)}$ (i.e. indexed by the partition that has $k$ parts of size 1, and one part of size $n-k$). The number of fixed points of an elements of cycle type $\mu$ is ${m_1(\mu) \choose k}$ (so this is the character). The multiplicity of $S^\lambda$ is given by the number of semi-standard Young tableaux of shape $\lambda$ and weight $(n-k, 1^k)$.
Case 4: unordered, distinct
We can identify $\Omega^k$ as the permutation module $M^{(n-k,k)}$. Although this has a basis that can be identified with the $k$-th exterior power of $\mathbb{C}^n$, that is not the correct representation because swapping two adjacent elements of a wedge monomial incurs a sign, while swapping to elements of an unordered set does not. This was investigated by Stier, Wellman, and Xu in Dihedral Sieving on Cluster Complexes; see Theorem 6.4. Similarly to Case 2, This gives a polynomial, which when evaluated at eigenvalues of a permutation matrix, gives the character. As for the multiplicities, they are given by the number of semi-standard Young tableaux of shape $\lambda$ and weight $(n-k,k)$.
