Why does a short circuit generate fire?
When a short circuit happens, provided the source impedance is null, the impedance is only determined by wires and is extremely low (let say resistance $R < 1\Omega$).
Therefore, by Ohm's law, if there is no current limitation, the current drawn is high (for example, with $V_\mathrm{rms} = 230\mathrm{V}$, $I_\mathrm{rms} > 230\mathrm{A}$). In this situation, the Joule effect is predominant and it follows $\mathcal{P} = UI = RI^2= \frac{V^{2}}{R} > 52\mathrm{kW}$.
It also happens if a really short time ($\tau < 0.01\mathrm{s}$): therefore it can be seen as an adiabatic transformation. This condition leads to a huge temperature difference ($\Delta T>1000\mathrm{K}$): copper wounds will quasi instantaneously melt and might vaporize, igniting everything around.
This is why circuit breakers have a magnetic safety: they break the circuit before the current reaches high values and destroys conductors (they are installation protections, they are not design to protect people. Only Residual Current Protection with low value can do so).
Given values are order of magnitude, to accurately determine it you will need models (or abacus) which rely on many factors (voltage source, source impedance, section and length of wires, heat coefficient transfer, etc.).
I would like to point out that, except in the movies, it isn't the case that "When a short circuit occurs it's obvious that there is fire".
Now it is true that, for most, "short circuit" evokes a picture of sparks and fire but, in fact, most short circuits simply result in a malfunctioning device, not sparks and fire.
A practical notion of "short circuit", in this context, is an unintended low resistance path for electric current. This low resistance path may (but not necessarily) result in much larger current though connected conductors and other circuit elements.
So, what do we need, in addition to this unintended low resistance path, for a fire?
Essentially, we need an electrical source that can deliver enough energy, before any protection elements open the circuit, to cause sufficient overheating in the attached circuit and/or source to start a fire.
The heating is simple to explain. The conductors in the circuit have non-zero resistance. Electric current through a resistance generates heat proportional to the square of the current
$$p = i^2 R$$
where $p$ is the power, $i$ is the current through and $R$ is the resistance. This is the reason that, for example, we can produce heat with electric heating elements as used in, e.g., electric ovens.
In addition, the electric source has internal resistance which means that the source will generate heat when there is a current through.
As to why current through a resistance produces heat, I'll quote the Wikipedia article "Joule heating":
Joule heating is caused by interactions between the moving particles that form the current (usually, but not always, electrons) and the atomic ions that make up the body of the conductor. Charged particles in an electric circuit are accelerated by an electric field but give up some of their kinetic energy each time they collide with an ion. The increase in the kinetic or vibrational energy of the ions manifests itself as heat and a rise in the temperature of the conductor.