The point "I" marks the transition from a low-current glow discharge, where the gas characteristics determine the behavior, to a high current arc, where electrode thermionic or high-field emission and even evaporation dominate.
Below is a "first-order" analysis of the transitions at points "D" and "I".
At low levels of current ("B" - "D"), what's happening here is that an electron emitted from the cathode collides with and ionizes gas atoms on its way to the anode, producing additional electrons, which can ionize additional atoms, etc. For $n_0$ initial electrons emitted by the cathode, with the cathode-anode spacing $d$, the number of electrons reaching the anode $n_a$ is:
$$ n_a = n_0 e^{\alpha d} $$
Here $\alpha$ is the first Townsend ionization coefficient; $1/\alpha$ is the average distance between ionizing collisions. On average, each emitted electron produces $ \left( e^{\alpha d} - 1 \right)$ new electrons.
It turns out that:
$$ \frac{\alpha}{p} = f \left( \frac{E}{p} \right) $$
where $p$ is the gas pressure, $E$ is the applied electric field, and $f$ denotes a functional relationship. (So at constant pressure, as the applied voltage increases, $\alpha$ does also.)
Of course, each ionized gas atom results in a positively charged ion as well as an electron; these ions drift back to the cathode. (Total current $I=I_0e^{\alpha d}$ is the sum of the electron and ion currents across any plane between the electrodes.)
What happens when these ions reach the cathode is crucial to the dark current / glow discharge transition. When ions collide with the cathode, some fraction $\gamma$ will "kick out" a new electron from the cathode; $\gamma$ is the second Townsend ionization coefficient. With this effect included, the total number of electrons emitted from the cathode $n_{0t}$ is:
$$ n_{0t} = n_0 + \gamma n_{0t} \left(e^{\alpha d} -1 \right) $$
Note the feedback effect now present: each electron emitted by the cathode can produce additional cathode electrons via the two effects. Solving:
$$ n_{0t} = n_0 \frac{e^{\alpha d}}{ 1 - \gamma \left(e^{\alpha d} -1 \right)} $$
$\gamma$ is also a (different) function of $E/p$:
$$\gamma=g (E/p)$$
From this formula one can identify the Townsend criterion for a spark breakdown, where the total number of electrons (and the current) go to infinity:
$$ \gamma \left( e^{\alpha d} -1 \right) = 1 $$
What happens next depends on the source of electricity. If we were gradually increasing the voltage of a "stiff" source (one with low series resistance) connected directly across the tube, the "infinite" current when the Townsend criterion is met, would run away directly to the arc region beyond "I", bypassing the glow discharge region entirely.
Suppose instead that we have a fixed voltage source connected to the tube via a large variable series resistor. (See this nice analysis, for example.) The operating point is varied by adjusting the resistor. Then, after breakdown at "D", the tube can operate stably within the glow discharge region, with the tube operating point automatically adjusting to the source (voltage supply + variable resistor). [The operating point is stable if the total resistance (variable resistor + incremental resistance of tube) is positive.]
The glow discharge current can be increased by reducing the external variable resistor. However, glow discharge current cannot be increased indefinitely. As the tube current and voltage increase, the bombardment of that current heats the electrodes, leading to increased thermionic emission from the cathode. With sufficient heating (i.e. at point "I"), cathode thermionic emission and material evaporation creates a lower-voltage "alternative" to the glow discharge, and a full-blown arc results. In this mode, the original gas in the tube is in a sense redundant; the arc can create its own plasma from vaporized electrode material.
[Note: there appears to still be uncertainty about the relative importance of thermionic and field emission in arc behavior.]
Electrons do "fill up your body" when you jump up and hit a high voltage wire - there is a property called the capacitance of the body that determines how much the voltage increases when you add a certain amount of charge - mathematically, $C = \frac{Q}{V}$.
But it's not charge that kills you, it is current: charge flowing per unit time. Since it takes relatively few electrons to bring the body up to 30,000 V or so, there is not much charge flowing and nobody gets killed. But you may have noticed a static "shock" when (especially in winter) you walked across a carpet, then touched a metal door and got a shock. As you walked across the carpet you built up static charge (with an associated potential that could reach several 10's of kV); and all that charge "leaks away" when you touch a grounded (conducting) surface. But while you can "feel" the current it's not enough to kill you.
So how much charge is there on your body when you are charged to 30,000 V? It's a bit hard to estimate the capacitance of a human body, so we'll use the physicist's trick of the "spherical cow": we approximate the human body as a sphere with 1 m diameter. The capacitance of a sphere is given by
$$C_{sphere} = 4\pi\epsilon_0 R = 0.11 nF$$
At 30 kV, that gives a charge of 3.3 µA; if that charge comes out of your body in 1 µs*, it would result in a peak current of 3.3 A which is why it feels like quite a jolt; however, the total amount of energy is only $\frac12 C V^2 = 0.05 J$ - and that is not enough to kill you. It's enough to kill sensitive electronic circuits, which is why you have to be careful how you handle "bare" electronics, especially in winter (low humidity = build up of static electricity as conductivity of air is lower).
EDIT
- if the current flows in 1 $\mu$s, that suggests that the time constant of body capacitance and skin resistance should be on that order. Since time constant is RC, solving for R gives about 10 kOhm. That’s a rather low resistance: skin resistance is higher, so peak current will be lower.
Best Answer
I think that you are mixing up two things one is an arc and the other is a Townsend discharge.
An electric arc is often formed by two electrodes coming together, creating a lot of heat and then the electrodes being separated to form a small gap in which the arc is formed.
The very high temperature created causes a plasma (ionised gas) to be formed between the electrodes.
The plasma is maintained by the thermionic emission (electrons "boiled off" from the very hot electrode) and field (electrons pulled out of electrode by a large electric field) emission of electrons from an electrode.
Those electrons traverse the gap where the atoms are colliding with one another and the electrons and causing excitation and ionisation of the atoms due to their very large "thermal" kinetic energy.
In another process (Townsend discharge) the mobile electrons which initiate the ionisation of the atoms are generated during the ionisation process itself.
Consider a gas which is between a positive and negative electrode.
Some electrons and ions are created in the gas by cosmic rays, ionising particles produced by radioactive decay etc.
The electrons (and ions) will be accelerated by the electric field between the two electrodes and in doing so gain kinetic energy.
These electrons will collide with the neutral atoms of the gas.
If the electric field is large enough (called the breakdown voltage and about $3\,\rm MV\,m^{-1}$ for air) then the energy gained by an electron between collisions with atoms will be sufficient to ionise an atom thus producing an ion and another electron which can then contribute to further ionisations.
So the number of electron and ions will multiply exponentially (cf avalanche) with the result that the gas now has a lot of mobile charge carriers within it - it has become a conductor.
Air as a conductor video.