Okay so first off.
The most important thing you should proably do to understand how orbits work, is to go buy and play Kerbal Space Program. Get yourself to a point where you can land on Duna and you should have a pretty good feel for how deltaV works. Even just a Mun landing would get you the basics.
Next, let’s talk about these numbers that are confusing you:
By far the most relevant number you need to look at is 11.2 km/s. This is the speed needed to escape Earth (aka “escape velocity” or the “second cosmic velocity”). If we subtract Earth’s orbital speed from this we have 11.2 - 7.9 = 3.3 km/s. Thus if we start at LEO, we would need around 3.3 km/s to leave the “Earth system”. If we go slower than this, we will not leave, but instead have a very long elliptical orbit which will eventually descend back toward Earth.
The moon’s orbit is pretty high up in Earth’s gravity well. So to get to the same altitude as the moon requires a very similar amount of deltaV. But if we want to stay at that altitude or pull into the orbit of the moon we’ll need to use a little more deltaV (so you might see velocities greater than 4 km/s to get to the orbit of the moon). For a fly-by mission, we do not stay in orbit so typically the second burn is not needed. Thus we have your deltaV to fly-by the moon of 3.1ish which is slightly-less-than Earth’s escape velocity.
Meanwhile if we had used exactly 3.3 km/s, we would come out of Earth’s sphere of influence and would pull into a solar orbit which looks very similar to Earth’s. In order to get to Mars, however, we would need to have a higher orbit. Thus we need just a little bit extra velocity, which is where you get your 3.6 km/s deltaV.
You also might see a smaller number to get to Mars such as your 3 km/s deltaV. This is probably because that mission started out in a higher orbit (with more energy), and thus required less energy to escape (i.e. they did not start at LEO as we have been assuming all this time).
Finally, there’s a number that’s confusing me:
Where are you getting 3.18 from? Oh, I think i see, you misplaced your decimal. You meant 31.8 km/s.
31.8 km/s is the “fourth cosmic velocity” in technical jargon. It’s the speed you would need to leave Earth and come to a standstill, cancelling out all of Earth’s orbital velocity (and then fall into the sun). If you want a full, math-filled discussion of this, then look at the google book: “Theory of Interplanetary Flights” By Grigor A. Gurzadyan (starting probably at chapter 6 on page 51).
The book even goes on to describe the next number 8.8 km/s. This is the “third cosmic velocity” (16.6 km/s: the velocity needed to escape from the sun when starting from LEO) minus the “first cosmic velocity” (7.9 km/s: the speed needed to orbit Earth at LEO). In other words, 8.8 km/s is the deltaV you would need to leave the solar system when starting from LEO.
The idea here being that you can get to the sun more easily by first leaving the sun then doing a small maneuver to cancel your remaining velocity and push you back toward the sun and letting its gravity pull you straight back.
That’s a very specific case of trying to get to the sun from Earth though; you don’t need to worry about any of that to understand deltaVs.
Hope that helps!