First lets give the box some dimensions: $w$ for width and $h$ for height.
A level, stationary box
The scenario you described above is almost correct if $F\,h\leq m\,g\,w$.
The net torque on the box would be $F\frac{h}2 - N\frac{w}2$, which if $N=m\,g$ would result in a net counter clockwise (negative in my chosen reference frame) torque. This would mean the corner not in red would be pushed into the floor, supporting some of the weight of the box, reducing $N$ until $F\frac{h}2 - N\frac{w}2 = 0$.
An accelerating box
If there is a net torque on the box (i.e. $F\frac{h}2 - N\frac{w}2 \geq 0$) then the box will be accelerating around the red corner. Note that to accelerate around the red corner would accelerate the center of mass. Since it's accelerating we can no longer claim that the net forces are zero. In particular now $F\gt F_\text{friction}$ and $N\gt m\,g$.
So Your first and second questions are answered by the fact that yes $N$ will increase as the box starts rotating, but that increase will allow it to overpower the weight of the box allowing the center of mass to accelerate upwards. Once the center of mass has moved upwards, there is now room for the corner to rotate without penetrating the ground.
As for your third question. No $F\neq F_\text{friction}$ once the box starts accelerating. $F_\text{friction}$ will reduce once the box starts rotating.
For your forth question, the net linear acceleration of the CG is not zero. You are correct that the CG describes an arc of a circle.
If you would like to calculate these values I would proceed as follows:
The moment of inertia of a box about its corner is
$$I=m\frac{w^2+h^2}3 \, .$$
The angular equivalent of $F=ma$ is
$$\tau=I\,\alpha=I\,\dot\omega=I\,\ddot\theta \, .$$
Now before we looked at the net torque about the center of gravity. While it's possible to solve this problem using that origin, choosing the red dot as our center allows us to skip a few steps:
\begin{align}
\tau &= F\frac{h\,\cos(\theta)+w\,\sin(\theta)}2-m\,g\frac{w\,\cos(\theta)-h\,\sin(\theta)}2 \\
\ddot\theta &= F\frac{h\,\cos(\theta)+w\,\sin(\theta)}{2I}-m\,g\frac{w\,\cos(\theta)-h\,\sin(\theta)}{2I} \, .
\end{align}
Unfortunately, this is equivalent to the large oscillation pendulum problem by a rotation of coordinate system. As such there is no analytic solution, but a numerical solution could get you $\theta(t)$.
Of course this solution would only be valid up to the point where friction would give way $F_\text{friction}>\mu_\text{static}N$.
The book is assuming that the ball is rolling without sliding, so the direction of rotation is fixed by that constraint. Also, if there is no sliding, the problem is completely time reversible. When you time reverse the forces, they point in the same direction as before, essentially because there is a t^2 in the accelerations, so the signs are unchanged. What static friction is doing here is simply transforming some of the translational kinetic energy into rotational kinetic energy when the ball is accelerating down the hill, and the opposite when the ball is decelerating up the hill. In that sense, static friction is always impeding what gravity is trying to do to the translational kinetic energy, but when the ball rolls uphill, static friction yields more translational kinetic energy than you would have had at that same height if you turned off the static friction. Surprisingly, this means that when a ball rolls toward an upward ramp, at will go higher up that ramp if the ramp's surface is rough than if the ramp's surface is perfectly smooth.
Best Answer
The forces acting in the $y$-direction, assuming no motion in that direction, are:
$$F_N=mg-T\sin\theta,$$
where $F_N$ is the Normal force exerted by the floor on the spool.
The forces acting in the $x$-direction are:
$$F_x=T\cos\theta-F_f,$$
where $F_x$ is an assumed net force acting in the $x$-direction and $F_f$ is a friction force. We usually model friction with a simple model:
$$F_f=\mu F_N,$$
with $\mu$ a friction coefficient.
Substituting we obtain:
$$F_x=T\cos\theta-\mu(mg-T\sin\theta),$$
$$F_x=T\cos\theta-\mu mg+\mu T\sin\theta.$$
a) Let's now take the particular case of $\mu=0$ (no friction at all), then:
$$F_x=T\cos\theta,$$
and:
$$F_f=0.$$
This means that if $\cos\theta > 0$, then $F_x > 0$ and there will be acceleration to the right ($x$-direction). Also, there will be (slippery) anti-clockwise rotation because there a torque $T$ acting about the central axis of the spool. Rotation is caused by angular acceleration $\dot{\omega}$, according to $TR_1=I \dot{\omega}$, with $I$ the inertial moment of the spool about its central axis.
b) In the case where $\mu >0$,
$$F_x=T\cos\theta-\mu mg+\mu T\sin\theta,$$
and $F_x >0$ only if:
$$T > \frac{\mu mg}{\cos\theta-\mu \sin\theta},$$
in which case acceleration in the $x$-direction will occur. But there will also be rotation because two torques act on the spool with resultant torque $M$:
$$M=F_f R_2-TR_1,$$
$$M=\mu (mg-T\sin\theta) R_2-TR_1$$
If $M>0$ then rotation will be clockwise, for:
$$\frac{\mu mg R_2}{\mu T \sin\theta-TR_1} >0.$$
c) Case of rolling without slipping:
If the friction coefficient $\mu$ is sufficiently large we have a regime of rolling without slipping which can be represented mathematically as:
$$v=\omega R_2$$.
In that case the spool would be accelerating to the left ($-x$-direction) and the spool will be rotating anti-clockwise.