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I know that $\pi_3(SO(4)) = \mathbb{Z} \oplus \mathbb{Z}$. We can choose an explicit identification as follows: given $(i, j) \in \mathbb{Z}$, we have a map $\phi: S^3 \to SO(4)$ which sends a unit quaternion $u$ to the map$$\phi(u): \mathbb{R}^4 \to \mathbb{R}^4, \text{ }\phi(u)(x) = u^i xu^j.$$

My question is, what is the crux/intuition behind all this?

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    $\begingroup$ If you are asking about Milnor's exotic sphere, then in the last chapter of Characteristic Classes by Milnor & Stasheff, there is an alternative construction. The main idea is by choosing such i and j, we are being able to control the Euler class and 1st Pontrjagin class of the vector bundle. We want to keep Euler class to be 1 so that when looking at sphere bundle we have a homotopy sphere(which is a homeomorphic sphere by Smale), and the extra room for the Pontrjagin class can change the Milnor invariant. $\endgroup$
    – Mingcong Zeng
    Commented Aug 12, 2015 at 14:11

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It probably helps if you notice that the unit quaternions are isomorphic to $SU(2)$, and the unit quaternions act as rotations on the quaternions by both left and right multiplication. So, we get a homomorphism

$$ p: SU(2) \times SU(2) \to SO(4) $$

It's easy to check that this is onto and that the kernel is $\pm (1,1)$. So, we get a double cover $p: SU(2) \times SU(2) \to SO(4)$, which therefore induces an isomorphism

$$ \pi_3(p) : \pi_3(SU(2) \times SU(2)) \to \pi_3(SO(4)) $$

Since $SU(2)$ is a 3-sphere, $\pi_3(SU(2)) \cong \mathbb{Z}$, so

$$ \pi_3(SO(4)) \cong \pi_3(SU(2) \times SU(2)) \cong \pi_3(SU(2)) \oplus \pi_3(SU(2)) \cong \mathbb{Z} \oplus \mathbb{Z} $$

But if you take the argument I gave and carefully follow all the details, you can get the explicit identification of $\pi_3(SO(4))$ with $\mathbb{Z} \oplus \mathbb{Z}$ described in the original question! For this it helps to notice that the map

$$ f_i : SU(2) \to SU(2) $$

given by

$$ f_i(u) = u^i $$

has degree $i$, so

$$ \pi_3(f_i) : \pi_3(SU(2)) \to \pi_3(SU(2)) $$

is multiplication by $i$.

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