How does the light source fire a single photon in the double-slit experiment

I think there's a bit of confusion here. The double-slit experiment was not performed with "single photons" - it's very hard to even consider what that would mean. At its heart, it is a thought experiment, and it's not really possible to make a real-life device that tests it.

The first low-intensity experiment (Taylor 1909) was challenging the EM field interpretation of photons - the idea was that if photons were localised concentrations of the EM field, as you lowered the intensity, there would be no photons to interfere with each other, and the diffraction pattern would disappear. When the experiment was low-intensity enough that Taylor couldn't distinguish between photons emitted and photons absorbed, he noted that the diffraction pattern still exists, so the photons couldn't just be localised concentrations of the EM field. Dirac had a different explanation - he considered that each individual photon was capable of interacting with itself.

Later, the experiments were repeated not with light, but rather, electrons. Electrons are a lot more convenient than photons, since they obey the Pauli principle: it makes a lot more sense to say "an electron here, an electron there". And you can emit individual electrons, which was first tested in 1974, and it was found that individual electrons do in fact display the same interference pattern. Later, it was found that the same pattern also appears for atoms and complex molecules (the current "world record" has the experiment done with a molecule with more than 800 atoms, at ~10 000 atomic weights; the experiment gets much harder with bigger "particles", since it requires much more precision). But we'll stick to electrons, since they're quite convenient.

Emission of individual electrons is still quite tricky, but they have a few important properties. They carry a charge, and they have mass. Both of these can be measured, and while this does disturb the electron (change its trajectory), it doesn't absorb it. Photons, on he other hand, will be absorbed by any measurement, which makes them tricky to deal with.

So you can measure with precision to individual electrons how many electrons were emitted from your emitor, and how many were absorbed on the detection surface. More importantly, you can try experimenting with what happens when you measure the electrons on the way between the emitor and the detector - and that's when real quantum phenomena come in.

Which path does the electron take through the double-slit apparatus? If you're already in the modern quantum physics mindset, the question doesn't really make any sense (and by modern, I still mean the first half of the 20th century :). But as we've seen, electrons can be measured on the way, so what if we put detectors on the two slits? Well, as was of little surprise to the people involved, the pattern disappears completely (this was predicted by Feynmann and others long before the actual experiment was done in 1961). What's more interesting (and still predicted in advance), putting the detectors behind the two slits also destroys the pattern.

But this is really a matter for another question entirely. At the core, the answer to your question is "the full experiment cannot be done with light, and was never attempted with light". The best that was attempted was to reduce the intensity of the incident light so low that the energy corresponded to one photon of a given energy - by adding distance and barriers in the light's path. But we know of no way to emit individual photons, or to detect them with certainty, or to measure their paths - and it may very well be outright impossible, not just infeasible. Now, this in itself is quite enough to discredit the dual theory of light (unless you go with Dirac's assertion), but the real experiment was done with single electrons, and continued with bigger and bigger quantum particles, up to the 800+-atom monster I mentioned earlier.

A light beam is not like a swarm of photons flying through space - the relationship between light beams and photons is rather more complicated than that. This is discussed in What is the relation between electromagnetic wave and photon? though a proper discussion requires going deeper into the subject of quantum optics that many of us consider sane.

Anyhow, the photon is basically the unit of energy exchange with the light beam i.e. if the power of the light beam is $W$ joules per second then the number of photons emitted per second is this power divided by the energy per photon:

$$ N = \frac{W}{h\nu} $$

The travel time of the light is $\tau = \ell/c$, where $\ell$ is the length of the light beam in your experiment, so to have only one photon at a time you need the number of photons per second to be of order $1/\tau$.

$$ N = \frac{W}{h\nu} = \frac{c}{\ell} $$

And solving for the power gives:

$$ W = \frac{h\nu c}{\ell} $$

So that's all you need to do. Make your light beam power equal to or less than $W$ calculated as above and you'll have only one photon present at a time.

You can use a single photon source, such that there is (provably) only one photon in the system at any one time. These are quantum mechanically single-photon emitters, as distinct from just turning down the intensity of a source emitting random photons.

One convenient example is the nitrogen-vacancy centre in diamond. A single NV centre only has one excitation and so can only emit a single photon each time it's excited.

Because the experiment is so inefficient (very few photons actually make it through the slits) finding a nice paper that directly shows it is tricky. However there are plenty of similar experiments using these sources (example) for duality experiments.