Recursion can be tricky even for experienced developers. I’ll try to explain to the best of my understanding how it works using the usual suspect: the Fibonacci sequence.
Fibonacci Sequence
In mathematics, the Fibonacci sequence is described as follow:
Fibonacci sequence is a sequence of numbers where each subsequent number is the sum of the previous two numbers in the sequence, except for the first two numbers, namely 0 and 1.
The sequence can be written like this:
$$ F_{n} = \begin{cases} 0 & \quad \text{, } n = 0 \\ 1 & \quad \text{, } n = 1 \\ F_{n1} + F_{n2} & \quad \text{, } n > 1 \end{cases} $$
The beginning of the sequence is thus:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144…
How it Translates in Programming
We can write this sequence in Ruby using a recurion, like the code shown below.
def fibonacci(n)
return 1 if n <= 2
fibonacci(n  1) + fibonacci(n  2)
end
The conundrum is on line three, when the method calls itself twice on the same line.
This is what is called recursion: a method or function which calls itself until a return condition is reached.
In the code above, the exit condition is on line two: if n <= 2
.
Let’s look at what happens when we n = 3
(\(F_{3}\)):
We know that \(F_{3} = F_{2} + F_{1}\) ; let’s see how it works:
$$ \begin{split} F_{3} = F_{2} + F_{1} \\ F_{3} = (F_{1} + F_{0}) + F_{1} \\ F_{3} = (1 + 0) + 1 \\ F_{3} = 2 \end{split} $$
Summary

The final return value of the requested Fibonacci number comes from the last line of our code:
fibonacci(n  1) + fibonacci(n  2)
. 
The line
return if n <= 2
is executed only during the recursion when \(F_{n} > 1\) is to be calculated. 
The method needs to know what the previous two answers (\(F_{n1}\) and \(F_{n2}\)) are before behing able to calculate \(F_{n}\). In other words, in every case the procedure works its way down to
n == 1
and then builds the final result with each return.  The Fibonacci joke is worse than the last two jokes you heard, combined.
See also Tail Recursion.