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.

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 and boundary map $\pi_{10} S^7 \cong \pi_{10} (Sp(2),Sp(1)) \overset{\partial}{\to} \pi_9 Sp(1)$. To understand this geometrically, I suppose one has to understand the proof of the isomorphism, which requires a lift $D^{10} \to Sp(2)$ lifting the map $D^{10} \to S^7$. I believe that it might be possible to describe this map fairly explicitly using coordinates, eg using a connection for the bundle.