Why is the horizontal force on sled $2\cdot F$ rather than just $1 \cdot F$? I was wondering why the net horizontal force on the sled was $2 \cdot F$ rather than just $1 \cdot F$. Is that the case because of the reactionary force of tension on the pulley? I think my understanding of tension is off.

 A: Yes. The tension in the string is $F$ and this acts twice on the sled/pulley because the string is wrapped around the pulley. So the net force on the sled/pulley is $2F$ to the right.
If you are drawing a free-body diagram for the pulley alone, remember that we are told that the pulley is massless. So the net force on the pulley must be zero, even though it is accelerating to the right and so is not in equilibrium. The string exerts a force $2F$ to the right on the pulley, so the sled must also exert a force $2F$ to the left on the pulley. And so the equal and opposite force that the pulley exerts on the sled is $2F$ to the right.
Another way to get the same result is to consider the motions of the mass $m$, the sled/pulley and the centre of mass of the sled/pulley/mass considered as a single system. Suppose the centre of mass of the whole system accelerates to the right with acceleration $a_0$; the sled accelerates to the right with acceleration $a_1$; and the mass accelerates to left with acceleration $a_2$. Then we have
$(M+m)a_0 = Ma_1 - ma_2$
But we know that $ma_2=F$ because there is a force $F$ acting to the left on the mass. We also know that $(M+m)a_0=F$ because the net force on the whole sled/pulley/mass system is $F$ to the right (when we consider the whole system, the tension in the string is an internal force so it cancels out and can be ignored). So we have
$F = Ma_1 - F \\ \Rightarrow Ma_1 = 2F$
which confirms that the net force acting on the sled/pulley is $2F$ to the right.
