Because of polarity, he is assuming that Vout is the highest potential, and is calculating the current going from Vout to Vin.
He is saying he wants to find the amount of voltage dropped across R1 by calculating the current of the circuit (by ohms law as the voltage of the circuit divided by the resistance of the circuit), and then multiplying by R1, and adding that to Vin to get V+.
He alternatively could have gotten the circuit current and multiplied by R2 and then subtracted that voltage from Vo to get V+.
Your voltage divider assumes that Vin is 0V (ground). This is currently not the case, though it is how the voltage divider method is introduced as it makes calculations slightly easier.
When you do a voltage divider, you are assuming that the current in the circuit is in series, and thus equal everywhere. You can then equate this current through each resistor, by ohms law, as a voltage divided by a resistance (assuming Vin as the highest potential and Vo as the lowest potential).
(Vin - V+) / R1 = (V+ - Vo) / R2 ... cross mulitply
(Vin - V+) * R2 = (V+ - Vo) * R1 ... expand
VinR2 - V+R2 = V+R1 - VoR1 ... group similar terms
VinR2 = V+R1 + V+R2 - VoR1 ... factorize
VinR2 = V+ * (R1 + R2) - VoR1 ... rearrange
V+ = VinR2 / (R1 + R2) - VoR1
Now of course when Vo (the lowest potential) is ground and thus 0V this term goes away and you are left with the familiar voltage divider equation:
V+ = VinR2 / (R1 + R2)
However when one end of the voltage divider is not 0V, as this lecturer has assumed (since there's little point putting 0V into an Op-Amp) you have to include the voltage in your calculations.
