The action of $S^1$ on $\mathbb C^k$ yields a rank $k$ complex vector bundle $V=ES^1\times_{S^1}\mathbb C^k$ on $BS^1$. You are interested in the cohomology of $S(V)=ES^1\times_{S^1}S^{2k-1}$, the total space of the unit sphere bundle of the vector bundle $V$. The Serre spectral sequence of the sphere bundle has only two nontrivial rows, so it boils down to a long exact sequence, the Gysin sequence
$$
\dots \to H^{p-2k}(BS^1)\to H^{p}(BS^1)\to H^p(S(V))\to \dots
$$
The map to $H^p(BS^1)$ (i.e. the only nontrivial differential in the spectral sequence) is given by multiplication by the Euler class of the vector bundle, an element of $H^{2k}(BS^1)$. Let's write the cohomology ring $H^\ast(BS^1)$ as $\mathbb Z[u]$ where $u\in H^2(BS^1)$ is the Euler class of the universal complex line bundle. So we need to know the integer $d$ such that the Euler class of $V$ is $du^k$.
The action of $S^1$ on $\mathbb C^k$ is the direct sum of $k$ actions on $\mathbb C$, so $V$ is the direct sum of $k$ complex vector bundles of rank $1$ (line bundles). The Euler class of a direct sum of bundles is the product of the Euler classes, so the problem of determining the differential is reduced to the case when $k$ is $1$. In that case if the action of $S^1$ on $\mathbb C$ is by $\lambda \mapsto \lambda^m$ then the class is $mu\in H^2(BS^1)$; the line bundle associated to the action is the $m$th tensor power of the universal line bundle, and Euler class gives an isomorphism from the group of isomorphism classes of complex line bundles on a space $B$ to $H^2(B)$.
Therefore the Euler class of $V$ is $m_1\dots m_ku^k$. So the number $d$ is $m_1\dots m_k$, and if the numbers $m_j$ are all different from zero then the cohomology ring $H^\ast(X)$ is the quotient of $H^\ast(BS^1)=\mathbb Z[u]$ by the ideal generated by $du^k$.
