Consider the fibration $K(\Bbb Z_2,1)\to \ast\to K(\Bbb Z_2,2)$. As we know from Serre (see Hatcher's SSAT), $H^*(K(\Bbb Z_2,n);\Bbb Z_2)$ is the polynomial ring on generators $\operatorname{Sq}^I(\iota_n)$ where $e(I)<n$ (this notation is standard and you can find it in Hatcher's book).
Now consider the spectral sequence associated to this fibration: we have $H^*K(\Bbb Z_2,1)=\Bbb Z_2[\iota_1]$ in the left column and $H^*K(\Bbb Z_2,2)$ on the bottom row has polynomial generators $\iota_2$, $\operatorname{Sq}^1\iota_2$, $\operatorname{Sq}^2\operatorname{Sq}^1\iota_2$, etc.
- $\iota_1$ transgresses to $\iota_2$, and since the transgression $\tau$ commutes with Steenrod squares, $\tau(\operatorname{Sq}^1\iota_1)=\operatorname{Sq}^1\tau(\iota_1)=\operatorname{Sq}^1\iota_2$.
- On the other hand, $\operatorname{Sq}^1(\iota_1)=\iota_1^2$, and $\tau(\iota_1^2)=d_3(\iota_1^2)=d_3(\iota_1)\iota_1+\iota_1 d_3(\iota_1)=0$ (clearly $d_3(\iota_1)=0$ for degree reasons).
The second observation must be wrong, but for the life of me I can't see why. More generally, the same logic shows that any decomposable element cannot be transgressive. This is obviously not true because, among other things, it would contradict Borel's result.
I feel like I am missing something obvious and thought I'd post here in case others have the same confusion.
As you say, $\iota_1$ transgresses to $\iota_2$, so on the $E_3$-page, there will be nobody in $\iota_1$'s place. Accordingly, $\operatorname{Sq}^1(\iota_1)$ on this page is not going to be anyone's square and the Leibniz rule does not apply.