If a nuclear weapon is donated on the moon it would launch debris into space. The debris would then fall to earth like asteroids.
This is incorrect, and reflects a common misconception about space.
I am not trying to ridicule you - this is a very common point of confusion shown by these XKCD submitters and the producers of GI Joe: Retaliation. But let's oversimplify it: Lunar astronauts are not dangling upside down like Australians. They do not hang from the underside of the moon by the soles of their shoes, where they could bend up towards their feet, grab a rock and stretch down towards the earth, and then drop the rock on the earth with catastrophic results. They are well planted on the moon, and chucking a rock, hitting a golf ball, blowing up a small bomb, or even firing a high-speed rifle (1000 m/s muzzle velocity) at the Earth high overhead would simply cause it to land on the moon.
Even when holding onto the ISS, throwing the rock down at the earth would (counterintuitively) not cause it to hit the Earth! It would simply change the shape of the orbit, rotating it about the Earth. Eventually, the rock would intersect your path again. Throwing it radially inward, you would need to either throw it incredibly fast so that the shape of the new orbit intersects the Earth before it gets 1/4 of the way around - which might take as little as 22 minutes. That projectile in GI Joe was not moving nearly fast enough to reach the Earth again in minutes - it appeared to be traveling prograde, meaning it would end up further from the Earth, but even going retrograde that launch speed would have required days or weeks to deorbit.
To get off the surface of the moon way, way up into low-Earth orbit requires about 5 km/s of Delta-V. From there it's downward onto the surface, requiring about 8 km/s. If you could direct the energy of your nuclear weapon into motion of a projectile - producing a big gun with a nuke instead of gunpowder - you could launch $m$ kg from the moon to low-Earth orbit with $K$ terajoules of energy according to:
$$
K = \frac{1}{2}mv^2
$$
where $m$ is in kg, $v$ is in meters (not kilometers) per second, and $K$ is in joules. 1 kg requires 12.5 MJ of energy.
And on reentry, you'd loose a lot of that energy. It would slow down and ablate in the atmosphere. You're going up by 5 km/s to LEO, then down from LEO by 5 km/s to get back to your original energy level, and then you've got an additional 3 km/s of energy which needs to make up for the losses of hoisting the nuke to the moon and all that atmospheric drag.
You're probably better off redirecting an asteroid from further out. Rather than being deep in the Moon's gravity well, you just start way above that of Earth. Speaking of gravity wells, that reminds me of another xkcd source which appears to show the Moon's gravity well as being much shallower than I interpreted it as based on these Delta-V numbers, which cite some inaccuracy. Check my work, please!