Op-Amp working principle
What you were told is not theoretically correct. You have to change the way o thinking what really happens. It is the feedback loop that can force IN-=IN+ , not IN-=IN+ that makes the loop work.

The condition IN-=IN+ des not happen magically.  

Start thinking that \$Vout=A_0(IN_+-IN_-)\$ where \$A_0=10^6\$ 

This is the IN\Out relation of an open circuit (not in a feedback loop) OPAMP

When you close the OPAMP in a negative feedback loop with resistance \$R_1,R_2 \$ this relation is still valid but the loop will change everithing

When you apply a voltage to the IN+ of the OPAMP, at time \$T=0^+\$ its output will  saturate to \$V^+\$(positive rail) as \$IN_+=Vin\$, \$IN_-=0\$ and so \$Vout=10^6(Vin-0)=too much\$.  

Then, thanks to the feedback loop \$IN_-\$ will go from \$0\$ to \$V_{out}*\frac{R1}{R_1+R_2}=V^+*\frac{R1}{R_1+R_2}\$. 

What happen next is that the value of \$V_{out}\$ will change as the value of \$(IN_+-IN_-)\$ chaged thanks to the loop. As a conseguange the value of \$IN_- \$ will change again and again Vout and \$IN_-\$ again and so on.....

At this point you should understand were we are going, as we have a negative feedback Vout and \$IN_-\$ will change until you reach a point of equilibrium. The point of equilibrium is when you have \$IN_-=IN+\$

This ideology can be reversed but be carful that in the other case the loop becomes a positive one and instead of reducing to  \$0\$ the difference \$IN_--IN+\$ this will increase untill you reach another equilibrium point that is in this case \$ V_{out}=V^+\$