Introduction

Seeing that you are showing more than a usual interest in an unfamiliar circuit, I decided to write this elaborate story for you.

Actually, this sophisticated circuit solution is not new. I first saw it in the Bulgarian translation of Clayton's Operational amplifiers book in the early 80s: Clayton, G. B. (1979). Operational amplifiers (2nd ed.). Newnes-Butterworths. Back then, I did not pay it much attention, but much later I saw it again in the even more mysterious circuits of negative impedance converters (NIC), and only then did I grasp the clever idea. Let's see what it is.

I will reveal the idea in successive steps, illustrating them with CircuitLab simulations. The schematics are conceptual; they contain elements with sample values.

Bridge viewpoint

Single-ended voltage divider

With two identical resistors R1 and R2 in series, we can make the most common voltage divider circuit. Its input voltage Vin and output voltage Vout are referenced to ground: Vout = Vin * R2/(R1 + R2).

^{simulate this circuit – Schematic created using CircuitLab}

Its attenuation is only 0.5. If we need to significantly reduce it, we will have to use a very large ratio between the resistances of the two resistors.

Differential voltage divider

But we can achieve the same result without significantly increasing the resistance ratio by adding a second voltage divider and subtracting its output voltage from the voltage of the first divider. Thus, the output voltage represents a small difference between two large, but closely valued, quantities: Vout = Vin * R2/(R1 + R2) - Vin * R4/(R3 + R4). In this way, we have actually invented the famous Wheatstone bridge, but used in a slightly less traditional way, where we vary its supply voltage as the input quantity. Thus, this "coupled divider" has a single-ended input and a differential output.

Non-inverting divider: R2/(R1 + R2) > R4/(R3 + R4). If we now slightly decrease the resistance R4...

^{simulate this circuit}

...the output differential voltage Vout will represent a very small portion of the input voltage Vin (significant attenuation), and its polarity will be positive. The circuit represents a positively unbalanced Wheatstone bridge or a non-inverting divider.

Zero-gain divider: R2/(R1 + R2) = R4/(R3 + R4). Then, if we make the resistance R4 equal to the others...

^{simulate this circuit}

... the output differential voltage Vout will be 0 V (infinite attenuation), regardless of the input voltage value. The circuit represents a balanced Wheatstone bridge.

Inverting divider: R2/(R1 + R2) < R4/(R3 + R4). Finally, if we slightly increase the resistance R4...

^{simulate this circuit}

...the output differential voltage Vout will represent a very small portion of the input voltage Vin (significant attenuation), but its polarity will be negative. The circuit represents a negatively unbalanced Wheatstone bridge or an inverting divider.

So, the attenuator implemented with a Wheatstone bridge has the property of passing the input voltage with the same polarity or inverted.

Feedback viewpoint

Now, let's apply all this wisdom in an op-amp circuit, where we force the op-amp to behave in a specific way by connecting through a voltage divider its output to one of the inputs (we introduce feedback). Also, we have to apply a (part of the) input voltage Vin to the input.

Positive feedback

If we connect the voltage divider to the non-inverting input, the feedback is "positive".

^{simulate this circuit}

As a result, the op-amp vigorously moves away from equilibrium, and eventually its output voltage reaches the supply rail.

Negative feedback

Conversely, if we connect its output to the inverting input, the feedback is "negative".

^{simulate this circuit}

The op-amp strives for equilibrium and eventually achieves it.

Mixed feedback

In the OP's circuit, something unusual is done – the op-amp output is connected through two dividers to both the non-inverting and the inverting inputs. This creates the bridge circuit from above, where the resulting differential voltage is the difference between the two divider's voltages. The feedback is "mixed".

Dominating positive feedback: If Schematic 1.2.1 is used (non-inverting divider), the resulting feedback is positive.

^{simulate this circuit}

Dominating negative feedback: If Schematic 1.2.3 is used (inverting divider), the resulting feedback is negative.

^{simulate this circuit}

Varying negative feedback:

The OP's circuit is even more sophisticated. In it, there is a small constant input voltage Vin. The air flow cools the heater (the resistor R4) and thus tries to decrease its resistance. The op-amp, however, senses this disturbance through its inverting input and starts to compensate for it. For this purpose, it increases its output voltage thus increasing the resistance R4 until it balances the bridge (the heater has a positive temperature coefficient, meaning that as its temperature increases, it increases its resistance).

^{simulate this circuit}

The interesting thing is that at the first moment when the power is switched on, the heater is cold; R4 is very low (< 1 kΩ), and the feedback is positive. When the filament heats up sufficiently, ​R4 exceeds 1 kΩ, the feedback becomes negative, and from that moment on, R4 = R2*R3/R1, accordingly its temperature, is maintained constantly. This provides some advantages in flow measurement which is not the subject of our consideration here.

The role of R2

This single-supply op-amp circuit must be biased, i.e., an input voltage must be applied. I used one of the possible (correct) ways by inserting the input voltage Vin into the R1-R2 divider.

Another (correct) way can be to inject a small current through a resistor into the input circuit, as shown in this question.

However, the method shown here, by applying current through resistor R2 to the op-amp output and the bridge input, is incorrect because the output behaves as a voltage source with very low resistance (I saw in one of your comments that you also noticed this and I support it).

Related answers

Here are links to three of my answers related to this topic:

Opamp circuits with feedback resistor connected to both inverting and non inverting terminal

Why can't the following configuration be considered negative feedback?

How a wheatstone bridge is used in operational amplifiers?