Well, suppose you have a flat bilayer of area $A$ that you bend into a sphere (vesicle). Because you want to conserve area [i.e. we ignore stretching and changes in thickness] the sphere radius is going to be given by $4\pi R^2=A$ i.e. $R=\sqrt{A/4\pi}$.
The bending energy is given by [using the notation of [2] in your answer]:
$$E_b = {K \over 2}\int dA (2/R)^2$$
where I used the fact that for a sphere the two main curvatures are the same and constant $K_1=K_2=1/R$. We can easily solve the integral (using the fact that the curvature is constant and $\int dA = 4\pi R^2$) and get
$$E_b = {K \over 2} 4\pi R^2 (2/R)^2 = 8\pi K$$
which implies, as you said, that the bending energy does not depend on the radius.
So, where's the catch?
Bending a patch $dA$ of membrane (where I call $dA$ the area element) does cost you more if you bend more, with an energy cost of $e_b \sim 1/R^2$. However, to make a curved vesicle (for which bending each "dA" costs you more) you need "less" building blocks (less patches $dA$) because the vesicle will be smaller. Indeed the "number" $N$ of $dA$s thay you have to use is
$$N=4\pi R^2/dA$$ i.e. it depends on the radius as $N\sim R^2$.
However, as you see, the two effects (bending more with less building blocks for a small vesicle or bending less with more building blocks for a big vesicle) compensate each other perfectly. The bending cost per area patch scales as $E_b\sim 1/R^2$, whereas the "number" of area patches scales as $N\sim R^2$, so that the cost of bending in this case is radius-independent because $$E_b\sim N e_b \sim R^2 * 1/R^2 \sim 1 $$
(..and by the way, this is only true for folding of 2D systems, as the bending energy always scales as $\sim 1/R^2$ but the number of elements scales as $\sim R^2$ only in 2D, in general as $\sim R^d$ in $d$ dimensions)
However you should not forget that this is valid only if :
you create the vesicle that has the same area as the patch you start with. If you try to create a bigger / smaller vesicle with the same amount of lipids, you will have to pay an additional "stretching" cost.
We are neglecting area stretching and thickness changes during our discussion.
we are neglecting the interface cost. Having the "sides" of your disk exposed costs you energy, as lipids would like to pack with each other.
Roughly, we can assume you pay an amount which is $E_i=\gamma P$ where $\gamma$ is a constant and $P$ is the perimeter. You only pay this cost when you have open ends. For a disk of area $A$ you have a radius $r=\sqrt{A/\pi}$ and thus
$$E_i=\gamma 2\pi r=\gamma 2\pi \sqrt{A/\pi}$$
However, for a sphere with the same area, which has no "open sides", $E_i=0$ but you pay the above mentioned cost of $8\pi K$ that you pay to form such a sphere. By summing up the two contributions with opposite signs, a total cost $E_s$ given by
$$E_s=8\pi K-\gamma 2\pi r $$ so that bending "into" a vesicle from a disk if favored only if the cost is as small as possible i.e. $E_s<0$ i.e. if
$$r>{4 K \over \gamma}$$
meaning that only big patch will like to turn into a vesicle because the energy gain involved in getting rid of the interfaces far outweighs the cost of being bent, whereas for small disks (which pay the same bending cost but have lower gain in getting rid of the interface) the transition is not favored.
This implies that, by area conservation, because $r=\sqrt{A/\pi}$
$$r=\sqrt{A/\pi}>{4 K \over \gamma}$$ and now using that for a sphere of the same area $A=4\pi R^2$ and combining, we get that formation of vesicles if favored if the final radius is
$$R>{2 K \over \gamma}$$
so that in the end only bigger vesicles are favored but that comes not from bending but rather from the interface cost associated with a disk.
Please however notice that here we are only considering the transition from disk to sphere, several other options might be more favorable than both.
