Can anyone explain to me in an intuitive way why that is?
It's not easy to say what you would consider intuitive and what not. I've made my guess and give an answer with no equations and with some statements I believe you could find intuitive. You'll have to pay a price, however. I'll make reference to several figures I have no time to draw and leave for you to reconstruct.
A slight change of notation. The force in case 1 I'll name $F_1$, the one in case 2 $F_2$. Other forces i'll define presently.
You agree that in case 1 the rod will simply accelerate with no
rotation, and its com will have acceleration $F_1/m$. Your question is about what will happen in case 2.
1) Add to case 2 two further forces: $F_1$ and its opposite (applied at the same point) which I'll call $F_3$. I ask you to agree that
addition of two opposite forces applied in the same place has no effect.
2) Consider the system of three forces as composed of force $F_1$
alone plus the couple formed by $F_2$ and $F_3$. I make a further assumption:
the effect on rod of several forces is the geometrical
(kinematical) composition of the separate effects of each one.
Since we know the effect of $F_1$ we are left to investigate the
effect of the couple $(F_2,F_3)$.
3) Consider another force $F_4$ equal to $F_3$ but applied to the
lower extreme of rod. I'm going to prove the following:
actions of couples $(F_2,F_3)$ and $(F_1,F_4)$ are identical.
To prove this consider another couple: $(F_3,F_5)$ where $F_5$ is
directly opposite to $F_4$. So $(F_3,F_5)$ is globally opposite to
Look at the system formed by 4 forces: $(F_2,F_3,F_3,F_5)$ (force
$F_3$ is doubled). It can be seen as formed by two subsystems:
$(F_2,F_5)$ and $(F_3,F_3)$.
The effect of $(F_3,F_3)$ is known: an acceleration leftwards of
magnitude $2F_3/m$. As to $(F_2,F_5)$ I ask you to accept another
two forces euqal in magnitude and direction, applied in different
points, have the same effect as a force of double magnitude applied in the midpoint.
So $(F_2,F_5)$ is equivalent to $2F_1$, which is opposite to $2F_3$. Then the total system $(F_2,F_3,F_3,F_5)$ has a null effect. But remember that it was formed of two couples: $(F_2,F_3)$ and $(F_3,F_5)$ and you see that both couples cancel each other. Since $(F_3,F_5)$ is opposite to $(F_1,F_4)$ we have shown that $(F_2,F_3)$ and $(F_1,F_4)$ are completely equivalent.
This is a non trivial result: we have shown that
translating a couple to a different position (leaving unaltered all parameters: magnitude and direction of forces and relative position of application points) leaves its effect unaltered.
4) Summarizig: $(F_2,F_3)$ and $(F_1,F_4)$ are equivalent. Then
applying both at the same time will have twice the effect of each couple alone. But applying both is the same as applying one couple of double arm: $(F_2,F_4)$. What will its effect be? I can appeal to symmetry to state
a couple of forces symmetrically placed wrt to com causes no motion of it, but only an angular rotation around it.
Then we may conclude that the same happens if only one of couples
$(F_2,F_3)$ and $(F_1,F_4)$ exists. Both keep com undisturbed causing an angular acceleration of rod around it. A not intuitive result!
5) And now we are at the end. We have seen that $F_2$ is equivalent to $F_1$ plus couple $(F_2,F_3)$. The former causes an acceleration of com, equal to $F_1/m$. The latter (the couple) has no effect on com but causes angular acceleration around it. Then this is the effect of $F_2$ alone, we were looking for.