If the moon were a uniform sphere, it would indeed appear dimmer at the edges of a full moon.
Surface roughness is the main reason the edges are not as dim as the sphere model predicts. When viewing a full moon from earth, the light is coming from nearly behind us. That means the edges of the full moon are illuminated by sun rays nearly parallel to the lunar surface. Now consider what on the moon those rays are going to hit. Just like here on earth at sunrise or sunset, vertical objects facing the sun will be better illuminated. When the surface has roughness, more of the glancing sunlight will hit more vertical parts of the surface.
What we are seeing at the edges of a full moon are predominantly the walls of craters, sides of rocks, and the like that are oriented towards the sun and therefore towards us. This more flat-on average orientation reflects significantly more light than a smooth-sphere surface would.
Imagine a very long cylinder. The circular part faces the Sun. The long part does not. Clearly, the long part of the area does not help the cylinder absorb extra sunlight from the Sun.
The reason you consider the Earth to be a disk while calculating the received flux is similar. I'll explain it in a different way which may be more intuitive. The Sun's light is spread across a full sphere of area $4\pi D^{2}$ where D = 1AU, the distance between the Sun and the Earth. The Earth casts a disk-shaped shadow that covers $\pi R^{2}$ area of that Sphere. The shadow covers some fraction of the Sun's light. The fraction of the Sun's total luminosity that Earth receives is then
$\frac{\pi R^{2}}{4\pi{D^{2}}}$
The third way to explain why it ends up being a disk, is somewhat less intuitive, but I'll include it regardless. The angle between the incident sunlight and the ground affects the power/(area of the ground). This is one of the reasons why the poles are cold whereas the equator is hot, and one of the factors behind the seasons. It turns out that taking the angle into account, the area that matters is just the component of the area that is perpendicular to the Sunlight (or the component of the normal vector of the area which is parallel to the Sunlight).
In short, the Earth receives energy from 1 direction (the Sun) and only shows the Sun a disk-worth of area, but the Sun is emitting energy in all directions over the area of a sphere
The part below does not directly apply to your question but I include it in case there's any confusion about multiple instances of spherical areas.
A confusing thing you will then find, is that the total spherical radius of the Earth is used again $4\pi R^{2}$ in these calculations. This is often because you might be interested in calculating the temperature of the Earth, by balancing Energy In and Energy Out.
Energy In will depend on the Disk Area of the Earth / Spherical Area of Earth's orbit and the Sun's Luminosity L: $L\left(\frac{\pi R^{2}}{4\pi D^{2}}\right)$
Energy Out will depend on the total spherical area of the Earth and its temperature, cooling according to the Stefan-Boltzmann Law. $\sigma T^{4} 4\pi{}R^{2}$
So we can balance the two:
$L\left(\frac{\pi R^{2}}{4\pi D^{2}}\right) = \sigma T^{4} 4\pi{}R^{2}$
In a sense, $\sigma T^{4}$ is the average flux (energy per area) received over the Earth.
Best Answer
The Earth receives approximately $6.8\text{mW/m}^2$ of reflected sunlight from the moon (see below for details of how I calculated that).
However, the sunlight is also absorbed by the moon and this raise the surface temperature. So the moon also emits thermal radiation towards the Earth (assuming the highest day time temperature of 400K, see comments below for more information), $\epsilon_{\text{moon}}(1-A)\sigma (400K)^4 = 89\text{mW/m}^2$
So the total power received from the moon (reflected + thermal) is 10,438 times weaker than sunlight, i.e.
$$ \frac{6.8\text{mW/m}^2 + 89\text{mW/m}^2}{1000\text{W/m}^2} = \frac{1}{10438} $$
To answer your question about how much that heats the Earth, let's assume that the average daytime temperature of the Earth is 20$^\circ$C and the average nighttime temperature is 10$^\circ$C (these estimates could be improved, but it doesn't really change the answer significantly).
Therefore the incident solar energy causes a temperature difference $\Delta T=10^\circ$C between night and day. So we know that 1000 $\text{W/m}^2$ (solar irradiance on the Earth surface) cause a temperature increase of around $10^\circ$C. Let's assume that moonlight will also cause a temperature difference but one that is scaled proportionally by its intensity. Moonlight is 10,438 (reflected and thermal energy) times weaker than sunlight, the change in temperature of the earth from absorbing moonlight is,
$$ \frac{10^\circ C}{10,438} = 958 \mu K $$
Good luck measuring that...
Assumptions and method
From 1 and 2 we know that $100\text{W/m}^2$ is reflected at the surface of the moon. From 3, let's multiply that by the solid angle subtended by the moon as viewed from the Earth as this will give us the amount of the reflected energy that hits the Earth. So, $100\text{W/m}^2 \times 6.5\times10^{-5} = 6.5\text{mW/m}^2$.