When I teach introductory op-amp analysis techniques, I emphasize the following to start with:
(1) Check for the presence of negative feedback. This means that the output of the op-amp is connected via some network to the inverting input of the op-amp.
(2) If negative feedback is present, assume the inverting and non-inverting input voltages are equal (not zero!).
In the case that negative feedback is present, the output of the (ideal) op-amp will be whatever it needs to be to make the inverting input voltage equal to the non-inverting input voltage.
To see this, assume that the non-inverting input voltage is made slightly (infinitesimally) more than the inverting input voltage. This should make the output more positive which will act to increase the inverting input voltage thus restoring the equality of the input voltages.
But what happens if we interchange the inputs? Rather than the inverting input, the output is connected to the non-inverting input instead. This is called positive feedback.
Now, as before, assume that the non-inverting input voltage is made slightly (infinitesimally) more than the inverting input voltage. As before, This should make the output more positive but, now, this will act to make the non-inverting input even more positive than the inverting input. The equality of the input voltages will not be restored, in fact, the non-inverting input voltage and output voltage will 'run away' - a real op-amp output will reach its maximum level and stay there.
So, despite the fact that a mathematical solution with equal input voltages exists in the case there is positive feedback, the solution isn't physically relevant since it is unstable - any disturbance will drive the circuit away from, rather than back to, the solution.
Another way to skin this poor cat!
Using Kirchhoff's voltage law (KVL) and Ohm's law you can write the following equation, where \$i\$ is the current in the mesh (clockwise direction):
\$
v_1 - R_1\,i - v_2 - R_2\,i = 0
\$
From this you get:
\$
i = \dfrac{v_1 - v_2}{R_1 + R_2}
\$
Apply Ohm's law to \$R_2\$ and you'll find the answer.
EDIT
(In response to OP editing his question)
You cannot apply equivalent circuit substitution that way. When you substitute a circuit section with an equivalent one, only the quantities external to that section are guaranteed to stay the same. When you substituted \$v_2/R_2\$ with its Norton equivalent, you lost track of the nodes across which you wanted to calculate the voltage (they disappeared inside the equivalent circuit).
If you want to go the route of Thevenin/Norton equivalence, you could substitute the series \$v_1, R_1, v_2\$ with its Thevenin equivalent: \$v_{Th}\$ (the open circuit voltage) in series with \$R_{Th}\$ (the resistance you see when you disable the voltage sources - i.e. substitute them with short circuits). You get then:
\$
v_{Th} = v_1 - v_2
\qquad
R_{Th} = R_1
\$
Note that \$R_2\$ wasn't touched in this substitution, thus the voltage across its terminals will remain the same. Therefore now you have \$v_{Th}\$ in series with \$R_1\$ and \$R_2\$, so you've got a voltage divider with \$v_{Th} = v_1 - v_2\$ total applied voltage, so you get:
\$
v_{R_2} = v_{Th} \dfrac{R_2}{R_{Th} + R_2} = (v_1 - v_2) \dfrac{R_2}{R_1+R_2}
\$
Best Answer
The arrow simply define the voltage direction notation. It's really super easy:
As you can see, that's necessary – it's not inherently clear which way around you'd measure voltage, and if you have an equation like \$y=-Ae\$, it'd be ambiguous.
Having that arrow is also a necessity, because what you do, introducing a ground potential at the positive input of the opamp, can't always be done – absolute voltages don't exist – everything is potential relative to an arbitrarily chosen point. And the arrow does nothing but define "the voltage I measure here is between the potential at the start of my arrow, and its tip – so subtract the start potential from the tip potential to get the voltage".