Planets that are larger and have higher gravity tend to have thicker atmospheres, since it is easier to hold onto gases.
Higher air pressure leads to a higher air density, and therefore, a greater the air resistance.
So, on a planet twice the mass of Earth with a proportionally thickened atmosphere, would a leaf fall faster or slower?
As I mentioned in a comment, I am making the following assumptions about this planet:
The mass of a sphere is its volume times its density: $$M = V*\rho$$ The volume is given by: $$V = \frac{4}{3} \pi R^3$$ By setting the mass equal to double the mass of Earth, we can find the radius: $$ \frac{4}{3} \pi R^3\rho_e = 2 \frac{4}{3} \pi R_e^3\rho_e$$ $$ R^3 = 2 R_e^3 $$ $$ R = \sqrt[3]{2} R_e $$
Surface gravity is calculated from the formula: $$ g = G \frac{M}{R^2} $$ Substituting from above, $$ g = G \frac{2 M_e}{(\sqrt[3]{2} R_e)^2} $$ $$ = \sqrt[3]{2} G \frac{M_e}{R_e^2} $$ $$ = \sqrt[3]{2} g_e $$
We decided that the mass of the air column was proportional to surface gravity:
$$ \sigma_{air} = \sigma_e \frac{g}{g_e} = \sqrt[3]{2} \sigma_e $$
Air pressure is the mass of the air column times the acceleration due to gravity (for a thin shell of atmosphere like Earth's, we can assume that the acceleration due to gravity is constant in the atmosphere without much error).
$$ P = g \sigma = \sqrt[3]{2} g_e \sqrt[3]{2} \sigma_e = \sqrt[3]{4} P_e $$
From the ideal gas law, we know that density is proportional to pressure at constant temperature (and composition):
$$ \rho_{air} = \sqrt[3]{4} \rho_{air,e} $$
The density of a leaf is much higher than air, so I will ignore buoyancy effects. I'll also assume a drag coefficient of 1, which strikes me as reasonable for a leaf.
The velocity of the falling leaf is when the drag force due to air resistance balances the force of gravity:
$$\frac{1}{2}\rho_{air} v^2 A_{leaf} = m_{leaf} g $$
Assume the leaf has some thickness $d$ and density $\rho_{leaf}$ which are the same on both planets. Then:
$$\frac{1}{2}\rho_{air} v^2 A_{leaf} = A_{leaf}d\rho_{leaf} g $$ $$ v^2 = 2 d \frac{\rho_{leaf}}{\rho_{air}} g $$ $$ = 2 d \frac{\rho_{leaf}}{\sqrt[3]{4} \rho_{air,e}} \sqrt[3]{2} g_e $$ $$ = \frac{1}{\sqrt[3]{2}} v_e^2 $$ $$ v = \frac{1}{\sqrt[6]{2}} v_e $$ $$ \approx 0.89 v_e $$
If we make the atmospheric increase larger, which I think is more realistic, then the leaf will fall even slower. If we increase the surface temperature or the proportion of light gases (Helium, Neon), which are also realistic, then the air density will be less and the leaf would fall faster. Making the planet denser (rock is not very compressible, so this would probably mean more Iron relative to Silicon) would increase the surface gravity, but since the atmospheric increase was proportional to surface gravity, this still makes the leaf fall slower.
It is worth noting that maintaining Earth's surface temperature in a thicker atmosphere implies that the planet orbits farther from its star or has a dimmer star.
tldr; For these assumptions, the leaf will fall slower on a larger planet.
The question has been edited to specifically ask about the buoyant force, so here's a little more info on that:
The buoyant force is given by the displacement of air by the leaf. The density of a fresh leaf, like other living tissues, is close to that of water, about $1000 kg/m^3$. The density of air at standard temperature and pressure is about $1.2 kg/m^3$. So the buoyant force is roungly $0.1\%$ of the force of gravity. If we double the mass of the planet and its atmosphere, then the density of the air increases to $\sqrt[3]{4} \times 1.2 kg/m^3$, or $1.9 kg/m^3$, so the buoyant force increases to almost $0.2\%$ of the force of gravity. This is still too small to an effect to bother including in the calculation.
The fact that a leaf falls slower in air than in a vacuum is almost entirely due to drag (aka air resistance), not buoyancy. You can test this by crumpling a leaf into a ball. It has the same buoyant force it always did, but much reduced drag, and it falls much more quickly.
Gravity is not the only thing to consider for atmospheric density; pressure (and by extension density but they are different properties) can also be increased by the amount of energy (read as heat) being stored in the atmosphere. Also, the relative strength of a magnetic field ( terrestrial or induced) can have an impact on how much of the atmosphere is retained as well.
This is best exemplified by Venus, with a very similar size and mass to earth but much denser atmosphere. There, bouyancy is your friend as it is going to be higher than on Earth but only needs to counteract the same gravitational force. Of course, building a bouyant airship that can survive that atmosphere without dramatically increasing weight is going to be a challenge, meaning that we're not comparing similar circumstances.
Finally, the other consideration is planetary density. Gravitational force increases proportionally to the square of your proximity to the center of mass, so a planet with the same mass as earth but (say) half the diameter is going to put you further down the gravity well yet may only have a similar atmospheric density. That would make the bouyancy have to work harder against gravity than on a planet with earth like density.
In short, there are a lot of factors to consider, gravity and atmospheric density being only two. All other things being equal however, I'd argue that if planetary density is equal, atmospheric density is proportional, etc. then the real deciding factor is whether the distance from the center of the planet is increasing at the square root of the increase in mass. If it's increasing at a lower rate, bouyancy will have a greater effect. If it's increasing at a higher rate, bouyancy will have less of an effect. But, it should be noted that this is a highly simplified way of looking at it and you would need a lot more specifics to figure it out on a case by case basis.
The question specifies "proportional". If we assume twice the gravity and air density, then: The answer is no. Terminal velocity would be neither faster, nor slower.
Note, the leaf would accelerate faster initially. But it would reach terminal velocity quickly. Terminal velocity is directly proportional to the sqrt of gravity and the sqrt of the inverse of the drag. So if "proportional" these factors cancel out.
Your primary premise is flawed.
Earth and Venus have comparable gravities but Venus has 90 times the atmosphere. Venus atmosphere
Other planets also differ widely. Saturn is huge but has close to Earth gravity, you can fit 1000 Earths into Jupiter but it has less than 3 times the gravity Planets gravity Neptune and Uranus are also close to Earth gravity, Uranus is a bit less than ours, yet both of these have atmospheres of perhaps hundreds of kilometres if not more.
To a first approximation, let's see what happens if we take the existing Earth and its atmosphere and dial up gravity to 2g.
But... terminal velocity occurs when drag equals weight. So we've doubled the drag but also doubled the weight - so the two cancel out and there is no difference.
