In the situation sketched in the question, only point 1. and 2. are relevant, as there is not any significant heat transfer between the bodies.
What coefficients should be in the first law according to the above assumptions:
There is no single/unique answer to this - you can choose the coefficients as you like.
The 1st law of thermodynamics says that
$$d U=Q-W$$
which is the same as you describe ($U$ is internal energy, $Q$ is heat, and $W$ is work). But, I could just as well have written it as
$$d U=Q+W$$
or with signs added in other ways. It depends on what I mean by $Q$ and $W$ in the specific case. To choose for example the version $d U=Q+W$, I have to state at the same time that $Q$ is heat absorbed and $W$ is work done on the body (by an external force). If I used $d U=Q-W$, then $W$ would have been work done by the system.
The method to remember it: Consider energy positive when it is entering (or absorbed by or added to) a body or system, and consider energy negative when it is leaving. Choosing this sign convention makes energy simple to add. If I lift a book up to a shelf, then I do work on the book (+W) by adding potential energy to the system. If I push a box across the floor, then I do work on it (+W) by adding kinetic energy to it. If on the other hand, the heavy box comes sliding towards me and hits me, so I am pushed away, then the box did work on me ($-W$). The work the box did on me is energy lost from itself; the amount of kinetic energy remaining in the box will necessarily be smaller after the collision.
Edit: I guess the second law might introduce a≤ (or is it ≥?) instead of an equal sign, since I talk about entropy.
The usual equation for entropy of a system is $S\geq\int\frac{dQ}{T}$. It states that entropy can never decrease in a process, if the whole isolated system is considered.
From comments
Work done ON A SYSTEM or BY A SYSTEM, I have never, ever seen a definition of. Work done by a force I have seen defined, on the other hand
You are correct, that only forces do work. The statement, "work done by a system" simply means that the system applies a force, which is doing work (on something else). In your meteor example, the meteor as the system is doing work on the planet by pulling in it toward itself through its gravitational force. (Note: The work done by the meteor on the planet is very, very small, since the displacement of the planet will be very, very small).
My guess is: work done by a system, is the thermodynamical force generated by the system multiplied by its conjugate variable on the surroundings. As in my example here: a gravitational forcefield is generated by the system on the surroundings, and the conjugate variable displacement on the surroundings is multiplied by it.
Yes, this is the mathematical definition, in short written as:
$$W=\vec F \cdot \vec s$$
or in general for non-constant forces:
$$W=\int \vec F \cdot d\vec s$$
where $\vec s$ is the displacement vector.
Consider a one dimensional infinite square well for simplicity. Treating the system quantum mechanically, the micro states are the eigenstates of the hamiltonian. These eigenstates are sinusiods, where the number of half wavelengths determines the energy level. The more wavelengths fit in the box (in other words, the shorter the wavelength), the faster the wavefunction phase oscillates. This increased oscillation with increased energy can be thought of as jiggling faster. So we see we have several different microstates with different energies.
Now if you were to heat the system up, then, as you said, particles would move from the low energy states to the high energy states, and so they would be jiggling faster. I think you understood this part already.
But now what happens if you compress the box? Then you would be making the wavelength of each eigenstate decrease, and so the frequency of oscillation of each eigenstate increases. Thus decreasing the box size also causes an increase in jiggling. However, where before each particle jiggled more because it moved to a higher energy state, we now have each particle jiggling more because it stays in the same state, but this state that it stays in starts jiggling faster.
I don't think you said anything wrong in your question, so I am not really sure where you are confused, but hopefully I cleared it up.
Best Answer
Suppose you have a thermodynamic system in a state $A$. In this state it has a certain amount of internal energy, $U_A$, because internal energy is a state variable. You can determine the internal energy by knowing only the state.
Now suppose the system undergoes some process - you don't know (or care) what - that leaves it in state $B$. Again, you can determine its internal energy by knowing only the state, and that energy will be $U_B$.
Clearly, to get from state $A$ to state $B$, the system had to gain a net amount of energy $\Delta U = U_B - U_A$. That fact is true no matter what the process was. Hopefully that's all clear so far.
But hopefully it's also clear that how that energy was transferred to the system depends on the process. There are many - or, well, at least several - different ways by which energy can be transferred in or out of a system, such as electromagnetic radiation, sound, gravity, or physically pushing on something. Different processes will use different methods, or different combinations of methods (because it's possible for a process to transfer some energy by EM radiation and some by sound, for example).
We've broadly grouped these methods of transferring energy into two categories, heat and work. Generally, the methods which involve the system pushing its environment around (or vice-versa) count as work, while others count as heat. (This should make sense because you need force and a change in position to have work.) So depending on the process by which the system gets from $A$ to $B$, the amount of energy transferred by heat methods and the amount of energy transferred by work methods can vary. Any basic thermodynamics textbook will give several examples to show how the distribution of energy transfer between heat and work depends on the process.
Of course, the total amount of energy transferred through all methods is always the same: it always has to be $\Delta U$. That's just energy conservation.