The point of the experiment is that in the aether, light along one arm of the interferometer will return at a different time than light along the other arm.
Forget about light for now and think of one of those moving walkways that you find in an airport (https://en.wikipedia.org/wiki/Moving_walkway). If I walk to one end of the walkway and back, that meaning I get a constant increase in speed going one way, and a constant decrease in speed coming back, will I have taken the same amount of time as if I didn't use the walkway at all? It turns out that the answer is no.
To see this, suppose the length of the walkway is $d$ and I walk with a speed $v$. If the walkway moves with a speed $v_w$, then I will be moving at a speed $v+v_w$ when I walk with the walkway, and a speed $v-v_w$ when I walk against it. The time it takes me to walk there and back is therefore $$\Delta t = \frac{d}{v+v_w} +\frac{d}{v-v_w}$$ as long as $v_w<v$ (which is has to be or I would never get back). This is in contrast to the time it would take with no walkway at all
$$\Delta t' = \frac{d}{v}+\frac{d}{v}$$
which is always less than the time it takes on the walkway whenever $v_w \neq 0$. So in fact, the average velocities are not the same as it takes me a longer time to walk the same distance on the walkway.
It is the same idea with the Michelson-Morley experiment. If the aether existed, the light ray that is parallel to the flow of the aether would experience a speed-up in one direction, and a slow-down in the other, which means it would take longer to return than the beam perpendicular to it. This time delay would cause the two beams to interfere when recombined, and this interference pattern can be viewed on a screen. Notice that the only solution to the equation $\Delta t = \Delta t'$ is when $v_w=0$, meaning that no interference pattern implies that there is no walkway, which in this case means no aether that changes the relative speed of light.