which would result in the piece of paper experiencing a force with magnitude of 2Ff in the opposite direciton of 2Ff.
It doesn't. I remember having this confusion too. Remember about internal forces or reactive/supporting forces for the entire system . Don't neglect drawing a free body diagram of the paper itself.
When you press a massless block against a table it doesn't experience double your applied from the table pushing back to support it. Imagine a horizontal force sensor on each wheel to measure the normal force: Intuitively (I hope) you don't add those numbers up, and they always read equal to each other.
UPDATE:
I realized the mistake I've made. I'm just confused as to how the net force on the paper is zero, yet its velocity increases (depending on the angular velocity of the wheel). There should be some work done on it and I struggle to understand where exactly it comes from.
Therein lies the distinction. Free body diagrams are a model to equate force to acceleration which is why they are drawn according to F=ma to relate force to the acceleration of objects. They don't determine internal forces because that's not what you set up to find.
So if you're interested in the acceleration of the paper then you need to actually draw a FBD of the paper istelf at some point. If you only draw FBDs of the wheels you are going to miss something.
When a free-body diagram says the net force along that axis is zero which means it does not accelerate. It does not mean the pressure/force experienced internally by the body is zero (internal forces).
The normal force applied to the paper in the horizontal direction by cancels out to zero so the paper has no horizontal acceleration. But the FBD for each roller has is own equation for the friction force, correspondingly, these equations also appear in the FBD of the paper. Note there are two of them, and they do not cancel out since the friction force of both wheels acts in the same direction even though the normal force behind the friction forces do not. That's why you need the FBD for the paper.
I assume the angular acceleration of the wheel and the acceleration at its center to be 0. Then applying
$\sum M_A = I_A \cdot \alpha_A$ yields $\ F_f = \frac{M}{r}$, where r is the radius of the wheel. However I have no idea whether I should repeat essentially the same process with the other wheel, which would result in the piece of paper experiencing a force with magnitude of $\ 2F_f $ in the opposite direciton of $\ 2F_f $.
