Aristoteles said, By the mean of a thing I denote a point equally distant from either extreme. Let's call that equidistant point y. Considering two data points y1 and y2, the equidistant point y would be found by setting the equality
y1−y = y−y2 and solving for y
y=y1+y2/2
which is the arithmetic mean of y1 and y2. Departing from this formula of mean being an equidistant point from the extremes, how do we come to the conclusion that the general formula for mean represent the average value, or most typical value of the data set.
Think probabilities and expectation. Let's say you have a probability distribution over the set $A = \{x, y\}$, i.e. $\mathbb{P}(x) = \mathbb{P}(y) = \frac{1}{2}$. Let $x, y$ be real numbers. The expectation of a random variable $X$ drawn from this particular distribution over $A$ is as follows $\mathbb{E}[X] = \frac{1}{2} \cdot x + \frac{1}{2}\cdot y = \frac{x + y}{2}$. Thus, you can say that this formula is arrived by placing a uniform distribution over the set of numbers.
Note that this generalizes readily. If $A = \{a_1, \ldots, a_n\}$, and you placed a probability mass of $\frac{1}{n}$ on each of the elements of $A$, you'd get that the expected value is $\frac{a_1 + \ldots + a_n}{n}$.
I want to repeat the main message of my answer. You are placing a uniform distribution over the elements. The uniform distribution says that each of the values is equally likely. So equally likely $\equiv$ most typical value. I hope this helps.