Why does regression to the mean work? According to Wikipedia, regression towards the mean is "the phenomenon that arises if a sample point of a random variable is extreme (nearly an outlier), in which case a future point is likely to be closer to the mean or average."
But wouldn't you expect any future point to likely "be close to the mean" regardless simply because that's the expectation value of the rv? What does this have to do with the fact that the previous observation was an outlier and why does this affect/increase the likelihood that a future point be close to the mean?

Let's do an example.
I'm going to flip a fair coin ten times. The expected number of heads is $5$. In statistics-speak:
$$
X \sim Binom(n=10, p=0.5)
$$
I do the flips and get the unusual result of $9$ heads and $1$ tail.
Now let's do it again. The expected number of heads is still five. That's regression to the mean.

But wouldn't you expect any future point to likely "be close to the mean" regardless simply because that's the expectation value of the rv? What does this have to do with the fact that the previous observation was an outlier and why does this affect/increase the likelihood that a future point be close to the mean?

This is the point of regression to the mean, and it's important to understand this to understand regression to the mean vs gambler's fallacy. The whole point of regression to the mean is that the expected result is the expected value. Got a particularly high score (e.g., $9$ heads) the first time? Your next attempt is likely to be lower...because it's always likely to be lower ($5$ heads). Got a particularly low score the first time? Your next attempt is like to be higher...because it's always likely to be higher.
In the gambler's fallacy, the assumption is that the expected value on the next attempt (or an attempt coming up soon) is influenced by previous results. In my experiment above with the ten flips of the coin, the expected number of heads is $5$. Got three in a row with $10$ heads? We might start to question that the coin is fair (reasonable), but if we believe that coin is fair, the expected number of flips is $5$ rather than $2$ or $0$ to offset the streak of heads.
