Why is $\pi_7(\mathbb S^4)=\mathbb Z \oplus \mathbb Z_{12}$?

I'm trying to visualize this fact, not prove it.

If we consider the (quaternionic) Hopf fibration $p:\mathbb S^7 \to \mathbb S^4$, where $\mathbb S^7$ is the unit sphere in $\mathbb H^2$ (we denote the quaternions by $\mathbb H$), the sphere $\mathbb S^4:=\mathbb H \mathbb P^1$ is the quaternionic projective line, and $p(z,w)=[z:w]$, then $p$ is an element of $\pi_7(\mathbb S^4)$ of infinite order. So, this map is responsible for the $\mathbb Z$ part of $\pi_7(\mathbb S^4)$.

Who is the element of order $12$?

Start with the quaternion Hopf fibration $$S^3\xrightarrow{i} S^7\xrightarrow{\nu}S^4.$$ The map $\nu$ is an element of Hopf invariant one so generates the infinite cyclic summand in $\pi_7S^4$. The map $i$ is the fibre inclusion. Now loop the fibration to get a homotopy fibration sequence $$\dots\rightarrow\Omega S^7\xrightarrow{\Omega\nu}\Omega S^4\xrightarrow\delta S^3\xrightarrow{i} S^7\rightarrow\dots$$ where $\delta$ is the fibration connecting map. Since $\pi_3S^7=0$ the map $i$ is null-homotopic. This implies that $\delta$ has a right homotopy inverse (ie. section) $\sigma:S^3\rightarrow \Omega S^4$. You can see easily using the adjunction $\pi_3\Omega S^4\cong\pi_4S^4$ that $\sigma$ is the suspension map (ie. the adjoint of the homeomorphism $\Sigma S^3\cong S^4$).

Now let $\mu$ denote the loop multiplication on $\Omega S^4$ and consider the composite $$\varphi:S^3\times \Omega S^7\xrightarrow{\sigma\times\Omega\nu}\Omega S^4\times\Omega S^4\xrightarrow\mu\Omega S^4.$$ It's not difficult to see that this map is a weak equivalence (recall that the loop addition on $\Omega S^4$ induces the addition on $\pi_*S^4$). Since both the spaces involved have CW homotopy type, $\varphi$ is a homotopy equivalence by the Whitehead Theorem.

In any case we get $$\pi_7S^4\cong\pi_6\Omega S^4\cong\pi_6(S^3\times\Omega S^7)\cong\pi_6S^3\oplus\pi_6\Omega S^7\cong \pi_6S^3\oplus \pi_7S^7.$$ From the way we defined $\varphi$ we see that the summand $\pi_7S^7\cong\mathbb{Z}$ corresponds cyclic summand in $\pi_7S^4$ generated by $\nu$. What's left is the summand $\pi_6S^3\cong \mathbb{Z}_{12}$, which is generated by a class normally denoted $\nu'$. From the above remarks we see that the corresponding summand in $\pi_7S^4$ is explicitly generated by the suspension $\Sigma \nu'$. This is who the element of order 12 is.

Now the class $\nu'\in\pi_6S^3$ itself actually has quite a nice description. Recall that $S^3$ can be identified as the unit sphere in the quaternions. The quaternionic multiplication induces a product of $S^3$ which turns it into a (nonabelian) Lie group. Consider the commutator map $$c':S^3\times S^3\rightarrow S^3,\qquad (x,y)\mapsto xyx^{-1}y^{-1}.$$ Note that this map is trivial when restricted to the wedge $S^3\vee S^3$. This means that it factors over the smash $S^3\wedge S^3=(S^3\times S^3)/(S^3\vee S^3)$ to give a map $$c:S^6\cong S^3\wedge S^3\rightarrow S^3.$$ I. James has shown that this map generates $\pi_6S^3\cong\mathbb{Z}_{12}$.