It seems that the resistance at the regulator input, which I shall call \$R_{REG}\$, is too low. Here's the circuit with some proper labels, and a more appropriate value for \$R_{REG}\$:
simulate this circuit – Schematic created using CircuitLab
The idea is to make the transistors become conductive when the 7812 regulator has reached its maximum current handling capacity, and let them take over current passing duties. At full pelt, these transistors are each passing nearly 5A each, current which must flow through their emitter resistors, which I will call \$R_E\$. Each emitter resistor \$R_E\$, then, would have a voltage across it:
$$ V_{RE} = I_{RE} \times R_E = 5A \times 0.1\Omega = 0.5V $$
To turn these transistors on enough to pass 5A of current, their base-emitter junction voltage \$V_{BE}\$ needs to be about 1V. This voltage is provided by \$R_{REG}\$ in series with the regulator. Whatever voltage appears across that resistor is shared between \$R_E\$ and the base-emitter junction. So, to have that transistor pass 5A, the voltage across \$R_{REG}\$ will have to be about:
$$ V_{RREG} = V_{RE} + V_{BE} = 0.5V + 1.0V = 1.5V $$
The voltage \$V_{RREG}\$ developed by resistor \$R_{REG}\$ is due to current flowing through it and into the regulator. That means we require \$V_{RREG}\$ to reach 1.5V (so the transistors are switched on), long before the regulator's current \$I_{REG}\$ reaches the maximum it can handle. Let's say that's \$I_{REG(MAX)} = 1A\$.
As it stands, with \$R_{REG} = 0.1\Omega\$, and regulator current maxed-out at \$I_{REG} = 1A\$, we can use Ohm's law to find \$V_{RREG}\$:
$$ V_{RREG} = I_{REG} \times R_{REG} = 1A \times 0.1\Omega = 0.1V $$
That's nowhere near enough to turn on those transistors. Regulator current would have to be many times that for the transistors to even begin to react. The 7812 will start throttling current long before that happens, and that's why you can't get even 2A out of your supply.
We need to calculate an appropriate value for \$R_{REG}\$ that will develop 1.5V when only 1A is passing through it and into the regulator:
$$ R_{REG} = \frac{V_{RREG}}{I_{REG}} = \frac{1.5V}{1A} = 1.5\Omega $$
If you replaced the 0.1Ω resistor \$R_{REG}\$ with 1.5Ω, then your problem is solved, but another one arises.
Power \$P_{REG}\$ dissipated in the regulator, at full load, is:
$$ P_{REG} = I_{REG} \times (V_{IN}-V_{OUT}) = 1A \times (17V - 12V) = 5W $$
Unless you have a beefy heatsink on that 7812, it will overheat and shutdown very quickly. To keep the regulator cool without needing a heat sink, we should aim for \$P_{REG}=1W\$, and the maximum current corresponding to that will be:
$$ I_{REG} = \frac{P_{REG}}{V_{IN}-V_{OUT}} = \frac{1W}{17V-12V} = 0.2A $$
That's our new target regulator current \$I_{REG}\$ that should develop the required 1.5V across \$R_{REG}\$. In other words, we require the transistors to switch on, and take over current passing duties, when regulator current reaches 0.2A:
$$ R_{REG} = \frac{V_{RREG}}{I_{REG}} = \frac{1.5V}{0.2A} = 7.5\Omega $$
Setting \$R_{REG}=7.5\Omega\$, or a little more, say 8.2Ω, will permit the transistors to switch on properly, and limit regulator current to about 0.2A, which it should be fine with.
These are ballpark figures, to be honest, since I haven't checked what \$V_{BE}\$ will actually be, when transistor collector current is 5A, and I haven't accounted for base currents. It's a starting point though.
