I have come to contradiction trying to compute $\pi_n(S^n \vee S^n)$ (n > 1). First of all we notice that the composition $S^n \to S^n \vee S^n \to S^n$ is identity map (the first arrow is embedding to the first sphere and the second arrow maps the second sphere to basepoint).
Applying the functor we observe that the map $\phi: \pi_n(S^n \vee S^n) \to \pi_n(S^n )$ has a section (in particular is epimorphism). This means the following sequence: $0 \to Ker(\phi) \to \pi_n(S^n \vee S^n) \to \pi_n(S^n ) \to 0 $ is exact and splits, i.e $\pi_n(S^n \vee S^n) = Ker(\phi) \oplus \pi_n(S^n )$.
Now the we need to compure the kernel but it is exactly those spheroids images of which lie in the second sphere, i.e. $\pi_n(S^n )$. Finally, we have $\pi_n(S^n \vee S^n) = \pi_n(S^n ) \oplus \pi_n(S^n )$.
But! If I did the same for $S^n \vee S^1$ I would obtain $\pi_n(S^n \vee S^1) =\mathbb{Z}$. However, from exact sequence for bundles we obtain $\pi_n(S^n \vee S^1) = \pi_n(S^n \vee S^n ... \vee S^n)$. Well, what is wrong?
This is wrong. The kernel of $\phi$ is just the kernel of $\phi$: those maps $f:S^n\to S^n\vee S^n$ such that their composition with the map $S^n\vee S^n\to S^n$ is nullhomotopic. This does not necessarily mean that the image of $f$ is contained in the second sphere, or even that $f$ is homotopic to a map whose image is contained in the second sphere (since the homotopy we have is only after composing with the map $S^n\vee S^n\to S^n$, not a homotopy of $f$ itself). If you don't believe me, I challenge you to write down a detailed proof that the image of $f$ must be contained in the second sphere.
(This description of the kernel of $\phi$ happens to be correct in this particular example, but for much more subtle reasons that do not generalize to other examples like $S^1\vee S^n$.)