A two-step Cohen iteration is a product

The check names $\check{q}$ for $q\in \mathrm{Add}(\omega ,1)$ occur in the constant name $\dot{Q}$, hence lie in F1's bounded carrier $R$. They form a dense suborder of the restricted iteration: for any $(p,\dot{q})$, the forcing membership clause gives a strengthening $p^{\mathrm{\prime }}\le p$ that forces $\dot{q}=\check{q}$ for some ground $q$, and $(p^{\mathrm{\prime }},\check{q})$ extends $(p,\dot{q})$. On this dense check-name suborder, $(p,\check{q})\le (p^{\mathrm{\prime }},{\check{q}}^{\mathrm{\prime }})$ exactly when $p\le p^{\mathrm{\prime }}$ and $q\le q^{\mathrm{\prime }}$. Send this pair to the finite function $r$ on $2\times \omega$ with $r(0,n)=p(n)$ and $r(1,n)=q(n)$. Restriction is the inverse on the dense suborder, proving forcing equivalence.

F2 factors the generic into the two one-coordinate generics; F3 says their union reconstitutes the full generic and either coordinate is Cohen-generic over the extension by the other.