From definition of Friction:
Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.
Now consider your problem. Let the force be $F$, mass of spool be $m$

If Friction is absent
If Friction were absent, then due to torque of $F$, the spool will rotate clockwise. Now, the bottom edge of spool may have a velocity in forward direction, or backward direction, given by:
$$\begin{align}\mathbb{v} &= v_{trans} -v_{rot} \\ \\
&=\frac{Ft}{m} - r\alpha t \\ \\
&= \frac{Ft}{m} - r\left(\frac{r\times F}{I}\right)t \\ \\
&= \frac{Ft}{m} - r\left(\frac{r F}{I}\right)t \\ \\
\end{align}$$
But we know that $\max(I) = mR^2$, so $\mathbb{v}$ above reduces to zero.
$$\begin{align}&= \frac{Ft}{m} - r\left(\frac{r F}{mR^2}\right)t \\ \\
&= 0
\end{align}$$
But we also do know that for most rigid bodies, $I < mR^2$.
From this, we have : $$v_{rot} > \frac{Ft}{m}$$
Therefore we can say in general, for rigid body, that:
$$\boxed{v_{rot} > v_{trans}}$$
Due to absence of friction, $\mathbb{v} \neq 0$ and also that the bottom point of wheel will have a net velocity in backward direction, and spool will slip.
If Friction is present
If Friction is present, it would make $\mathbb{v} = 0$, since it tends to remove relative motion between the bottom of spool and ground.
Now, the direction of friction will be decided by the velocity which was greater in the earlier case.
As we saw earlier, $v_{rot} > v_{trans}$ for a rigid body, $\mathbb{v}$ came out to be negative, hence friction will try to make $\mathbb{v}$ zero, or will be exerted in positive $x$ direction on spool.
Friction always opposes relative motion. So on ground, friction will act in -x direction (by third law of Newton).