$A\subset\mathbb R^n$ is measurable if for all $B\subset \mathbb R^n$
$$m^*(B)=m^*(B\cap A)+m^*(B\setminus A)$$
This is just a definition, and it basically says that the outer measure can be partitioned into the outer measure of the intersection of set $A$ with any other set, and the outer measure of the intersection of that set with the complement of $A.$
But why is it defined as such?
One thing it achieves is countable additivity: given two disjoint Lebesgue measurable sets $E$ and $F:$
$$\begin{align}m^*(E\cap F)&= m^*\left((E\cap F) \cap E \right ) + m^*\left((E\cap F) \cap E^c \right )\\[2ex] &=m^*(E )+ m^*(F) \end{align}$$
And relatedly, how does it connect with the alternative definition:
For all $\epsilon>0$ there exist an open set $G$ and a closed set $F$ such that $F\subset A\subset G$ and $m^*(G\setminus F)<\epsilon$ ?
Lebesgue's outer measure $m^*$ is constructed as an attempt to extend the notion of length of an interval, in a coherent way, to all of $\mathcal P(\mathbb R)$.
But then Vitali sets exist, which show that $m^*$ is cannot be a measure. Caratheodory's criterion gives you a family of subsets $\mathcal M$ that satisfies three things:
restricted to $\mathcal M$, the outer measure $m^*$ is a measure
$\mathcal M$ is a $\sigma$-algebra
$\mathcal M$ contains all open sets, and thus the Borel $\sigma$-algebra
The three points together give us a big enough domain (bigger than the Borel $\sigma$-algebra) where the idea of $m^*$ works and actually provides a measure. e: the Lebesgue measure.
On the question about how Caratheodory came up with such an idea, I cannot really say. With the criterion already in place, one may think that the motivation is to take sets where $m^*$ is additive.