The point is that $\eta\circ k=\eta+\eta+\dots+\eta$ is the $k$-fold sum of $\eta$ in $\pi_3S^2$. In particular $\eta\circ(-1)=-\eta$. Now,
$$k\circ\eta=\eta\circ k^2$$
and
$$\Sigma(k\circ\eta)=\Sigma k\circ\Sigma\eta=\Sigma\eta\circ\Sigma k=\Sigma(\eta\circ k)$$
by the commutativity of these elements after suspension. Hence
$$\Sigma(\eta\circ k)=\Sigma(\eta\circ k^2).$$
With $k=-1$ we have
$$\Sigma(\eta\circ(-1))=\Sigma(-\eta)=-\Sigma\eta$$
on the left, and
$$\Sigma(\eta\circ(-1)^2)=\Sigma(\eta\circ 1)=\Sigma\eta$$
on the right. In particular
$$\Sigma\eta=-\Sigma\eta,$$
showing that this element has order no greater than $2$.
Thus to show that $\Sigma\eta$ has order $2$ it will suffice to show that it is nontrivial. One way to accomplish this is by noting that the mapping cone construction commutes with suspension. In particular, the mapping cone of $\Sigma\eta$ is $\Sigma\mathbb{C}P^2$. If $\Sigma\eta$ were null-homotopic, then $\Sigma\mathbb{C}P^2$ would be homotopy equivalent to $S^3\vee S^5$ (which is the mapping cone of the constant map $S^4\rightarrow S^3$). That $\Sigma\mathbb{C}P^2$ does not split as a wedge of spheres is a consequence of the non-trivial Steenrod square
$$Sq^2:H^3(\Sigma\mathbb{C}P^2;\mathbb{Z}/2)\rightarrow H^5(\Sigma\mathbb{C}P^2;\mathbb{Z}/2).$$
If this is unfamiliar, then let us stick to the EHP sequence. We know that the suspension $\Sigma:\pi_3S^2\rightarrow\pi_4S^3$ is surjective by Freudenthal's Theorem. Since $\eta$ generates $\pi_3S^2\cong\mathbb{Z}$, and $\Sigma\eta$ has order $\leq2$, the group $\pi_4S^3$ is either $\mathbb{Z}/2$ or is trivial. Both of these groups are devoid of odd torsion, so to answer our problem it will suffice to work 2-locally.
Thus introduce the EHP sequence,
$$\dots\rightarrow\pi_5S^5\xrightarrow{P}\pi_3S^2\xrightarrow\Sigma\pi_4S^3\rightarrow 0$$
which becomes exact after localisation at $2$. By exactness, the right-hand group is trivial if and only if $P$ is surjective. But it is well-known that
$$P(\iota_5)=[\iota_2,\iota_2],$$
where $[\iota_2,\iota_2]$ is the Whitehead product of $\iota_2$ with itself. I calculated the valued of this Whitehead product here (at least up to sign), showing that
$$[\iota_2,\iota_2]=\pm2\eta$$
(the exact sign will depend on your particular conventions).
The point is that $P$ is not surjective, so it must be that $\pi_4S^3\cong\mathbb{Z}/2$, generated by $\Sigma\eta$.
Best Answer
Let $H(f)$ denote the Hopf invariant of $f\colon S^{2n-1}\to S^n$. Then $H(\tau\circ\eta)=(\text{deg}\,\tau)^2 H(\eta)$ and $H(\eta\circ\sigma) = (\text{deg}\,\sigma)H(\eta)$. Since $(-1)^2=+1$, there is no contradiction.
See Problem 14 in Milnor's Topology from a Differentiable Viewpoint or Exercise 1 on p. 428 of Hatcher.
Regarding your argument, why should $[\tau\circ\eta]=[\eta]\in \pi_3(S^2)$? What you say up to "That implies that ..." is all correct.