I tried to the find the homology groups the following way. I'm not sure if it is correct.
By excision, $H_m(S^n, \{x_1,x_2,...,x_k\})\cong \widetilde H_m (S^n/F),$ where $F=\{x_1,...,x_k\}$.
$\color{red}{S^n/F\simeq S^n\vee \overbrace{(S^1\vee S^1\vee...\vee S^1)}^{k-1 \text{ copies}}}$. .This is because of the following image:
So we have $\widetilde H_m(S^n/F)=\widetilde H_m(S^n)\oplus \widetilde H_m(S^1)^{k-1}$
It follows that $\widetilde H_0(S^n/F)=0\implies \widetilde H_0(S^n/F)=Z, H_1(S^n/F)=Z^{k-1}, H_n(S^n/F)=Z$
$ H_m(S^n/F)=0, m\ne 0,1,n$.
Is this correct, in particular the red colored part? Thanks