How does the physical motion of atom lead to photon emission? It's known that what we call a temperature is in fact molecular motion at microscopic scale. But at which point the emission of photons happens due to this physical motion, so that we can talk about the thermal emission?
the Wikipedia page on thermal radiation says the motion of atoms "..results in charge-acceleration and/or dipole oscillation which produces electromagnetic radiation..", what is beyond the terminology I am competent of.
May I ask for some explanation that doesn't require a degree in Physics to understand it? 
 A: When charge accelerates, it radiates energy. The "prototypical scenario" illustating this is the uniformly accelerated point charge and the Larmor Formula (see Wikipedia page of this name) that quantifies the radiated power (both the non-relativistic and relativistic versions are given on the Wiki page). For an everyday example, you can think of a dipole antenna, wherein charge undergoes roughly simple harmonic motion back and forth in the antenna's conductor, and thus radiates power (even a receiver antenna radiates a reaction field when its charge is accelerated by incoming radiation, which is how the TV detector man in the UK can get you if you don't have a licence: their detectors seek the reaction field from a receiver set).
So now, in a hot gas, we have a great deal of clumps of charge (gas molecules) bumping into one another. Each collision is naturally a violent acceleration for these charges, and so they radiate. Analogous processes happen in hot solids, although each molecule is more like the dipole antenna - heat represents vibrational energy in the lattice of molecules and there aren't really collisions as such.
In contrast, in deep space, hot gasses expand swiftly and pretty soon they become so rarefied and thin that there are very few collisions. Thus, in deep space there can be hot, "dark" gasses which do not give off their telltale radiation. Before anyone asks me whether this is a candidate for dark matter, I'll just say I don't know, I'm out of my depth and you need to ask another question for a real astronomer to answer.
