Ohm’s Law for voltage divider

This answer is probably inherently displeasing to the feeling of natural order for some :-) :

A law of nature is simply a statement of observed results under defined conditions.

Ohms law is essentially a statement that the ratio of the two two variables V & I is typically observed to remain approximately constant as the variables vary.

It is arguably saying the opposite of what it may seem - ie not so much
"R is the ratio between ..." but
it is more "if the ratio between V & I is constant then we call this constant resistance" and "this approximates typical behaviour of a significant proportion of real world products".

At any given moment R IS the ratio between V and I. If this ratio has changed then R has changed. So V/I never changes for specific values of V & I with all other conditions held constant, whereas dV/dI typically does change in the real world.

So R = V/I is an accurate statement
R = dV/dI is usually an approximation and
where it all falls apart it just means that the observation does not apply under thos e conditions.

That's woolier than I'd like but seems to convey what I'm trying to say. I hope :-).

The numerator is the 15K resistor because you are measuring the voltage across R2 -- it is assumed the positive probe of the voltmeter is at "Voltage Out", and the negative probe of the meter (not shown) is at the bottom of R2 (the - side of the 9V battery).

The current through R2 is set up by ohm's law:

\$I = \frac{V}{R}\$, so \$I = \frac{9v}{12K + 15K}\$ = 333 µA

So the voltage across R2, using \$V = IR, V = 333 µA * 15K = 5v\$

Combining the two equations, you get \$V = \frac{9v}{12K + 15K} * 15K\$

which is the same as the equation in the last line of your question: \$V = \frac{15K}{12K + 15K}*9v\$