Elliptic Bootstrapping for Gauge Transformations

I agree that you can't get a bound in $W^{3,3}$ for the reason you give: that multiplication doesn't exist (certainly $1 \in W^{2,4}$, so $1 \otimes W^{2,2} \to W^{2,3}$ would be the identity, but as you say $W^{2,2}$ does not sit inside $W^{2,3}$). I don't have a copy of Morgan's book available, but if you did have a $W^{3,3}$ bound I guess you would want to conclude using the compactness of the inclusion $W^{3,3} \hookrightarrow W^{3,2}$. I tried for a little while but do not see how to fix this argument.

Freed-Uhlenbeck prove the same theorem for connections on a principal $G$-bundle and connections in $W^{\ell, 2}$, $\ell \geq 2$ as their A.5 in "Instantons and Four-Manifolds". They only use the conclusion that $\|\tau_n\|_{W^{3,2}}$ is uniformly bounded, which you outline the proof of in your answer (there is, after all, a product $W^{2,4} \otimes W^{2,2} \to W^{2,2}$).

They conclude by noting that the compactness of inclusions implies that a subsequence $\tau_n'$ converges in eg $W^{2,3}$; then using the Sobolev multiplication $W^{2,3} \otimes W^{2,2} \to W^{2,2}$ you see using your equation that $d\tau_n'$ converges in $W^{2,2}$. This is enough to conclude that $\tau_n'$ converges in $W^{3,2}$, as desired.