$\DeclareMathOperator\id{id}$Let $\Omega^2(S^2,x)$ be the set of basepoint-preserving maps $S^2\to S^2$ (with basepoint $x=(1,0,0)$). Instead of taking the constant map $S^2\to S^2$ as the basepoint of $\Omega^2(S^2)$, let's take the identity map $\id_{S^2}$ to be the basepoint.

Certainly, it is the case that $\pi_3(S^2,x)\cong \pi_1(\Omega^2(S^2),\id_{S^2})$ since loop spaces have homotopy equivalent path components. Also, if $X$ is the path component of $\id_{S^2}$ in $\Omega^2(S^2)$, then $\pi_1(\Omega^2(S^2),\id_{S^2})\cong H_1(X)$ by the Hurewicz map. Hence, $\pi_1(\Omega^2(S^2),\id_{S^2})=H_1(X)$ is infinite cyclic.

Let $f:S^1\to \Omega^2(S^2)$ be the loop based at $\id_{S^2}$, which rotates $S^2$ once around the $x$-axis. This map is a low-dimensional case of the $J$-homomorphism and its homotopy class apparently generates $\pi_1(\Omega^2(S^2),\id_{S^2})$.

The map $f$ itself is very simple and highly geometric but the connection to the Hopf map gets a bit muddled. What is a direct, "elementary" proof of why the homotopy class of $f$ generates $\pi_1(\Omega^2(S^2),\id_{S^2})$? By "elementary," I mean some argument using elementary tools from homology theory, geometry, or homotopy theory that perhaps connects back to the Hopf map (or not) and could be understood by someone who does not know general results characterizing the image of the $J$-homomorphism.