For $i = 1$ you can argue using the fact that $S_n \to O(n)$ induces an isomorphism on abelianizations. $\pi_i^{st}$ are the homotopy groups of the group completion of $\coprod_n BS_n$, and at least for $i$ positive $KO_i$ are the homotopy groups of the group completion of $\coprod_n BO_n$.
For $i = 2$, you can argue that the map is surjective using Pontrjagin-Thom construction and the Atiyah invariant. $\pi_2^{st}$ is the set of cobordism classes of framed surfaces, and the map to $KO_2 = \mathbf{Z}/2$ factors through the "Atiyah invariant'' of spin surfaces, I'm not sure whether that boils down to something elementary.
For $i = 4,12,20,\dots$ the map is zero because the domain of the map is torsion.