Electronic – Maximum current vs temperature

If you look at the Ebers -Moll model, one can see that:

$$I_C = \dfrac{\beta_F}{\beta_F +1}I_E$$

which can be rewritten as

$$\beta_F +1 = \beta_F \dfrac{I_E}{I_C}$$

$$1+\dfrac{1}{\beta_F}=\dfrac{I_E}{I_C}$$

$$ \beta_F = \dfrac{1}{\frac{I_E}{I_C}-1}$$

Where

$$I_E = I_{ES}\{e^{\frac{V_{BE}}{V_T}}-1\}$$

where $$V_T=\dfrac{k_BT}{q}$$ is the thermal voltage.

Here it should be clear that the temperature cause a change in the thermal voltage which leads to a change in beta.

The equation is basically a statement about the distribution of energy per unit of charge, as the temperature increases, the exponential in the emitter current equation decreases, which causes the ratio of collector current to emitter current to increase, and since that ratio is in the denominator of the equation for beta, it has an inverse relationship, which causes beta to increase. So as temperature increases, beta increases.

The LM317 is used to provide a trim voltage - it's output will be around 1.25V or whatever it's internal reference voltage is, then the resistor/pot is used to derive a trim range of something like 0-140mV.
The PNP transistor to increase the current capability (the LM35 cannot provide 20mA by itself)

If we look at the numbers, we want a 16mA increase for 100°C. So this is 0.16mA per °C. We know the output of the LM35 is 10mV°C, so we can calculate the current through the 62.5Ω resistor:

10mV / 62.5Ω = 0.16mA, just what we need as the LM35 outputs 10mV/°C, which produces 0.16mA thorugh the resistor. Since the range is from 4-20mA, we need 4mA drawn to start with though. The LM317 will sink around 2.8mA (1.25V / 452Ω) so we just need another 1.2mA. 1.2mA * 62.5Ω = 75mV, so we trim the pot to about halfway for out zero °C reference at 4mA.

Then the voltage/current across/through the resistor increases. The LM317 current will stay static, so once trimmed it's just the output and PNP sinking the rest of the current. I'm not sure exactly how the LM35 is desinged internally, but it probably involves a compensated feedback amplifier stage to provide a low impedance output.
Roughly, the LM35 will sink current through it's supply pin, some of which comes through the base of the transistor and turns it on, until it's voltage at the output is the desired point.