Earlier answers seem to be off-topic, since their authours talk about real photons, while the question is asked about the picture of virtual photons which serve as interaction mediators for the electromagnetic interaction, even in the electrostatic case.
The most important thing to settle beforehand is that the picture of interaction by exchanging virtual particles assumes essentially quantum arrangement. That means some different way of thinking about reality and processes. I advice the popular book
Feynman. QED, The Strange Theory of Light and Matter
that explains it the best way. Here I tell very roughly the most important points.
In classical physics you can think that processes proceed as they are described as the time goes by. In quantum physics, you think by the following pattern 1:
- You imagine the process as a whole, from its initial state to its final state.
- For this process as a whole, you calculate a complex number called the probability amplitude (it is just a word, don't think of its sense). In the Feynman's book it is called 'arrow' for simplicity.
- You imagine all other possible processes that give exactly the same final state. For them you repeat steps 1 and 2. Sometimes you can skip very complicated processes because they give very small numbers.
- You add all probability amplitudes, and only after that you decide, whether the process has taken place at all.
For the interaction by the exchange of virtual particles, this means that the absorber has the same importance as the emitter. It is the presence of the absorber "in the right place and time" that makes the whole process possible at all.
Added later: Also for the words emitting and absorbing, they are used in some figurative sense, since the virtual photons are emitted and absorbed within the bounds of a single quantum process, and cannot reach some detector, for example. Also, the temporal order of interactions can switch depending on the viewpoint, so the roles of the emitter and the ovserver can switch as well. More about that in Feynman's.
Now we are ready to go through your questions.
What determines the energy and direction of the emitted photon?
The positions and velocities of both the emitter and the absorber. After some perplexed 4-dimensional algebra, that comes down to the usual Coulomb and Biot-Savart laws.
Notice that for two static charges the energy of the photon would be 0! Such photons would transfer only momentum, until at least one of charges would gain some speed. That corresponds to the fact that the Coulomb force does not produce power if the charge does not move.
How often can a particle emit a photon?
As often as it needed to make the interaction of needed strength, for given emitter and absorber.
How often can a particle absorb a photon?
As often as it needed to make the interaction of needed strength, for given emitter and absorber.
These two questions lead to the question "how often two particles actually exchange photons?" That is calculated by the value of action of the whole process (which you have considered on the step 1). Very roughly, you can take the energy of interaction $E$, the interval of time $\Delta t$, and then the action per that time would be $A=E\,\Delta t$. This action can be attibuted (very roughly) to the interchange of $n$ photons where $A=2\pi\hbar n=hn$ and $2\pi\hbar=h$ is the Plank's constant.
You see that the closer charges are, the more photons they exchange, and as the time goes by, more and more photons run between them. For macroscopic charges and distances, the number of photons would be very large, so the interaction feels smooth as the classical physics tells. For elementary particles flying by, it is not unusual to exchange only one photon (or none at all), which is one of the most interesting processes for the particle physics.
Can one particle emit and absorb multiple photons at once?
In the Quantum Electrodynamics (QED), no. In some other interactions, it is sometimes possible, for example, one gluon (which is charged with color charge) can emit two other gluons at once.
Where does the energy to emit a photon come from?
From the energy of the charged particle. But remember, the energy of a photon can be 0 (see above). So it is not needed to have some spare energy to take part in interactions. And sometimes the charged particle can get energy, if the other charged particle gives it.
Is the destination of the photon somehow pre-determined or is the photon simply emitted in the hopes of being absorbed?
The destination is determined: it is the absorber. But it is not pre-determined in some temporal sense, because the quantum process happens as a whole, and not by some consequtive stages. Just when the absorber happens to be there to catch the photon, it is emitted.
If the absorber does not happen to be there, actually the photons are emitted anyway. But that is a very special case: all these photons go back to emitter. They do not take any energy or momentum, and their very existence would be unobservable, but they show themselves in some subtler phenomena, being known as radiative corrections.
1 It is important to note that quantum physics can be represented in several ways mathematically equivalent. Here I tell only the Feynman Path Integral picture, which is the most natural for the story about virtual particles. But some explanations would sound wrong and would turn on the different side, if one would start with Schrodinger picture, for example.
