A generator may be deduced indirectly as follows (using only information available in Hatcher's Algebraic Topology (AT)):
One has a fiber bundle $$Sp(n-1)\to Sp(n)\to S^{4n-1}$$ (p. 383, Example 4.55 of AT ). Taking $n=2$, we have a fiber bundle $$S^3=\mathbb{H}^1=Sp(1)\to Sp(2) \to S^7.$$
From the long exact sequence of homotopy groups (Thm. 4.41 AT), we have $$\cdots \to \pi_{10}(S^7)\to \pi_9 Sp(1)\to \pi_9 Sp(2) \to \cdots$$ which is exact at $\pi_9 Sp(1)$.
Again from the fiber bundles $$Sp(2)\to Sp(3)\to S^{11}, Sp(3)\to Sp(4)\to S^{15}, \ldots$$ and long exact sequence, we see that we're in the stable range, so $$\pi_9 Sp(2)=\pi_9 Sp(3) = \cdots = \pi_9 Sp(\infty) = \pi_5 O(\infty)= 0$$ by Bott periodicity (p. 384 Hatcher).
So we have a surjection $$\mathbb{Z}_{24} =\pi_{10} S^7\twoheadrightarrow \pi_9 S^3.$$
By the Freudenthal suspension theorem (p. 360, Cor. 4.24 Hatcher), and from the table on p. 339 of AT, $$\mathbb{Z} + \mathbb{Z}_{12} \cong \pi_7 S^4 \twoheadrightarrow \pi_8 S^5 \cong \pi_9 S^6 \cong \pi_{10} S^7 \cong \mathbb{Z}_{24}.$$
Moreover, the $\mathbb{Z}$ factor of $\pi_7 S^4$ is generated by the Hopf map $\nu$ (Example 4.46, 1st paragraph p. 385 AT) associated to the quaternions $$S^3 \to S^7 \overset{\nu}{\to} \mathbb{HP}^1=S^4$$ Hence $\langle \nu \rangle \cong \mathbb{Z} \leq \pi_7 S^4$ must surject $\mathbb{Z}_{24}=\pi_8 S^5$ under the suspension map.
Summarizing, we see that $$ \pi_7 S^4 \geq \langle \nu\rangle \twoheadrightarrow \pi_{10} S^7 \twoheadrightarrow \pi_9 S^3,$$ where the first surjection comes from triple suspension, and the last map is the connecting map in the fibration exact sequence.
The connecting map is induced from the isomorphism $\pi_{10} S^7 \cong \pi_{10} (Sp(2),Sp(1)) \overset{\partial}{\to} \pi_9 Sp(1)$. To understand this geometrically, I suppose one has to understand the proof of Theorem 4.41 of Hatcher, but I believe that it might be possible to describe this map fairly explicitly using coordinates.