4-velocity and 4-acceleration in instantaneous rest frames

OK I think I see where your confusion lies.

You're talking about the four velocity and acceleration in the instantaneous rest frame $F'$, and as you say in this frame the four velocity is $(1, 0, 0, 0)$. Your mistake is to assume the four velocity is constant in $F'$, because it is not. Remember that after an infinitesimal time $dt$ the rocket is not longer in $F'$ - it is in a new instantaneous rest frame $F''$. The rocket's velocity in the new rest frame $F''$ is still $(1, 0, 0, 0)$, but in old $F'$ frame it has now changed due to the acceleration. Hence $d{\bf u}/dt$ in $F'$ is not zero.

If you're interested chapter 6 of Gravitation by Misner, Thorne and Wheeler derives the equations of motion that you started with.