You may find that you'll develop perspectives and views on circuits that work best for you. For some, the purely theoretical view is clearest. For others, they need to see how all the gears and wheels work to get it clear in their head.
So when an op-amp is used with feedback, a simple principle appears: the op-amp will drive its output to whatever voltage is necessary to make its inputs be equal.
This principle applies to correctly-designed feedback circuits, such as you have, and within practical supply voltage limits and a good few other things. But it can simplify one's view of it when thumbing through a circuit.
Looking at your circuit...
The ideal op-amp, on its own, will subtract the Vi- voltage from the Vi+ voltage and multiple the result by its huge open-loop gain. Odds on, this will make the op-amp output go as high as its positive supply or as low as its positive supply. It can't go beyond them, no more volts available.
With the feedback resistors added, it's all a bit more laid-back. The op-amp still does exactly the same thing but the overall circuit sees a more limited behaviour from it.
If Vin is driven with 2 V, the op-amp will do (0-2)*huge = -huge and its Vo output will take a swift trip towards its negative supply rail, trying to get to Vo = -huge V.
However, the div2 potential divider of R1:R2 is connected across Vin and the op-amp's Vo...and the op-amp's Vi- input is actually driven by that potential divider. So Vi- will be driven with the voltage halfway between Vin and Vo, as per simple potential divider behaviour. And as the op-amp's is driving towards its negative supply, Vo will soon travel down to -2 V on its way.
Now we have Vo = -2V driving the bottom of the potential divider and Vin = 2 V driving the top end of it. So the voltage out of the div2 is 0 V.
If the op-amp output carried on swinging negative to, say, Vo = -3 V, the R1:R2 potential divider would present the op-amp's Vi- input with ( ((2 - -3)/2) + -3) = -0.5 V. The op-amp would do it's gain thing and conclude (0- -0.5)*huge = +huge and start driving towards the positive rail. On the way, it would cross -2 V and start driving to -huge again.
But it doesn't go too far in your circuit, it stops nicely when Vi+ = Vi-.
As you know that your Vi+ is tied to 0 V, you also know that the op-amp will do its best in this circuit to keep its Vi- input at the same voltage. Hence the name 'virtual earth': it isn't one, but the op-amp will do it's best to keep Vi- pulled to that by the feedback network of R1 and R2.
From the op-amp's point of view, Rload is outside of the feedback divider so it just gets driven to Vo all the time.
All the stuff about being able to take the R1 current as Vin/R1, that comes from the knowledge that the op-amp output will pull the R1:R2 potential divider about until Vi- is at the same voltage level as Vi+. So it lets you simplify your calculations for Ir1 a little.