There is a way to explain it that's similar to what you said about multiplication by $n$.
Let $G$ be a Lie group, and let $f_1$ and $f_2$ be any two clutching functions that describe $G$-bundles $E_1$ and $E_2$ on $S^n$. Suppose that $f_1$ and $f_2$ agree at a base point of $S^{n-1}$. Let $c$ be some characteristic class of $G$-bundles of degree $n$; it could even be a cup product of standard classes such as Chern or Pontryagin or whatever. Let $f_3$ be the combination of $f_1$ and $f_2$ on the one-point union $S^{n-1} \vee S^{n-1}$. It is the clutching function of a bundle $E_3$ on the suspension $\Sigma(S^{n-1} \vee S^{n-1})$, which is the union of two $n$-spheres along an interval and thus homotopy equivalent to $S^n \vee S^n$.
Whether you define $c$ the old-fashioned way by obstructions, or the more modern way by pullbacks from classifying spaces, it is easy to argue that $c(E_3) = c(E_1) \oplus c(E_2)$. I.e., it's just the ordered pair of the characteristic classes of its parts. Now addition in $\pi_n$ is modeled by a map $S^n \to S^n \vee S^n$, and the induced map on $H^n$ takes $a \oplus b$ to $a+b$. (Your generalized question about $H^k(S^n)$ is of course non-trivial only when $k=n$.)
A shorter, more modern summary of the same story is as follows. The $X$ is a space and $LX$ is its loop space, then $\pi_{n-1}(LX) \cong \pi_n(X)$. The loop space of the classifying space $B_G$ is homotopy equivalent to $G$ itself, so $\pi_{n-1}(G) \cong \pi_n(B_G)$. A characteristic class of degree $n$ is any cohomology class in $H^n(B_G)$. The linearity that Milnor uses is the transposed form of the fact that the Hurewicz homomorphism $\pi_n(B_G) \to H_n(B_G)$ is linear. (If you expand this it out more explicitly, it isn't really different from what I say above.)
