Calculate the force between the earth and the bowling ball. It'll be G • (m(earth) • m(bowling ball)) / (r = distance between both mass centers)²

Simplify. You're getting g • m(bowling ball).

Now do the same for the feather. Again, the result is g • m(feather).

Both times you end up with an acceleration of g. If you want to put it that way: The force between the earth and the bowling ball is m(bowling ball)/m(feather) times as high as the force between the earth and the feather, but the second mass also is m(bowling ball)/m(feather) times as high, resulting in the same acceleration g.

Higher force on same mass results in stronger acceleration. Same force on higher mass results in lower acceleration. Higher force on equally higher mass results on equally high acceleration.

I just asked my professor this exact thing (if the ball would get to the earth sooner because it accelerates the earth towards it) like two weeks ago and my previous message + this message was his explanation.

PS: If you're looking at this from outside, the ball travels less distance before touching the ground (since the ground is slightly nearer due to pulling the earth more towards it), but also accelerates slower while accelerating the earth faster towards it. The feather gets accelerated faster towards the earth and travels a longer distance before touching the ground but doesn't accelerate the earth as fast towards it.

But because we're not outside, we only care about the total acceleration (of the earth towards the object and the object towards the earth), and that's g. We don't notice if (fictional numbers) the earth travels 1m and the object travels 1m or if the earth stays in place and the object travels 2m, what matters for us is how long it takes an object 2m away from the earth to be 0m away from the earth.