How does this spool have an angular acceleration opposite to its acceleration? Consider this spool. It rests on a rough horizontal force. The green lines are the ground, and a line parallel to the ground. The spool is pulled without sliding by the thread (with tension T), with force applied on the inner radius at an angle theta to the horizontal.

My question is, we can see that the force T would have it rotating counter clockwise, but the force is only applied to the right, so the whole figure would move right, opposite to rotation! How is that possible, since its not slipping?
 A: 
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 direction ($x$-direction) and the spool will be rotating clockwise.
A: The force at T would have the spool rotating counter-clockwise, but that's not the only force on the spool.  In steady state, a equal and opposite force from T must also be applied by the ground.  This force would have the spool rotating clockwise.
However, it's better to look at torques than forces.  The torque from the string is T x R1, while the torqe from the ground is -T x R2.  The latter torque is obviously larger.  If you pull on the the string extending level at T, the net torque will be clockwise, and the spool will roll to the right.
It may help to think of the limiting case where R1 is zero.  Or put another way, the string is attached to the center.  Of course pulling the string rolls the spool towards you.
If you could arrange the spool rolling on a shelf or something so that the part where the string inwinds from extends below the shelf (R1 > R2), then the spool would roll away from you as you pulled the string.  When R1 = R2 there is no net torque, and the spool doesn't rotate at all, until the force exceeds the friction and it slips.
