Op - Amp: Linearity, Negative Feedback and Virtual Short
The most simple way for explaining the negative feedback effect is to split the whole procedure in several steps:

Example: Non-inverting opamp stage with two equal resistors in the negative feedback loop. Supply voltages +/- 10volts. Open-loop gain Aol.

Step 1: Apply a DC input of +1V at the non-inv. terminal.

Step 2: Because each amplifier has a certain signal delay the feedback is not yet active at t=0 and the output will "jump" to the supply rail (+10V).

Step 3: Now we have 10/2=5 volts at the inv. terminal (and still +1V at the non-inv. terminal). Hence, the voltage difference Vd between both opamp input terminals now is negative (-4.5 volts).

Step 4: Now, the opamp output "wants" to go to the negative rail (-10V). However, on the way from +10V to -10V the output voltage crosses the linear transfer region and finds one value which - together with the corresponding feedback signal - exactly fulfills the equilibrium point which is defined as Vout=Aol*Vd.  

Step 5: This point of equilibrium within the linear transfer region ist stable (it is the DC operating point) because the negative feedback causes a kind of correction as described as follows:  

When (during the mentioned crossing effect) the output voltage gets a bit too large (overshoot) , the feedback voltage at the inv. input also is too large and the difference Vd becomes smaller - thereby reducing the ouput voltage again (and vice versa).     

Example: Aol=100. Feedback factor k=0.5. The real closed-loop gain is 

Acl=100/(1+0.5*100)=1.9608.

and Vout=+1V*1.9608=1.9608V.

Voltage difference Vd=1-1.9608*0.5  and Vout=Vd*Aol=100(1-1.9608*0.5)=1.9608 volts (q.e.d.)

Comment: From the calculation it is evident that for larger open-loop gains Aol the voltage difference Vd becomes much smaller - and thus can nearly fulfill the common approximation Vd=0 (virtual short).