A Tseitin conversion preserves satisfiability

Set $(x,y,z)=(1,0,0)$. Then $F=(1\cdot 1)\vee 0=1$. Its computed extension is $(u,v,w)=(1,1,1)$. The nine clause values in order are $1,1,1,1,1,1,1,1,1$. More generally, for any original assignment with $F=1$, setting $u=1-y$, $v=xu$, $w=\max (v,z)=1$ satisfies the gate clauses and final unit.

Conversely the first two clauses force $u=1-y$. The next three force $v=xu$, and the following three force $w=\max (v,z)$. The last clause forces $w=1$, whence $F=1$. For example $(x,y,z)=(0,1,0)$ has forced gate values $(0,0,0)$, so the last clause fails and there is no satisfying extension.

The satisfying original assignment from step 1.1 with the inconsistent extension $(u,v,w)=(0,0,0)$ still makes $F$ true, since $F$ does not mention fresh variables, but makes both $u\vee y$ and the unit $w$ false. Thus the enlarged formula is not equivalent to $F$ on arbitrary assignments to all six variables. Its existential projection onto the original variables is exactly $F$, by the two preceding arguments. No choice is required: the valid extension is computed uniquely.