There have been some good answers already given but I want to note another aspect, namely a heuristic involving the Mobius function. Let $\mu(n)$ be the Mobius function. The Riemann Hypothesis is equivalent to the claim that for any $\epsilon >0$ one has that $$\sum_{1 \leq k \leq x}\mu(n) = O(x^{1/2+\epsilon}).$$ This equivalence stems from making an explicit integral for $1/\zeta(s)$ in terms of the Mobius function which converges up the the 1/2 line if one has the above bound on the sum.

Now, it is reasonable to guess/hope/assume that $\mu(n)$ is essentially random in the sense that the non-zero values behave essentially like a fair coin with heads corresponding to 1 and tails corresponding to -1. It turns out that if one has a fair coin, and one keeps flipping it, then the difference between the number of heads and tails after $x$ flips will with probability 1 be $O(x^{1/2+\epsilon})$. So if the Mobius function behaves like a random coin flip, or even close to a random coin flip we should expect RH to hold with probability 1.

Note that this isn't the only thing satisfying about this framing of the Riemann Hypothesis. This sort of shows one major reason that RH is important: a lot of sieves and other ways to get a handle on the primes involve inclusion/exclusion arguments, which is essentially what $\mu(n)$ is for. So in a moral sense RH says that if one is doing inclusion/exclusion with primes, one cannot get very large deviations in how much at any given stage one needs to include or exclude.

(Edit to add: actually this is essentially the same sort of approach as noted in the link by Pace above.)

There have been some good answers already given but I want to note another aspect, namely a heuristic involving the Möbius function. Let $\mu(n)$ be the Möbius function. The Riemann Hypothesis is equivalent to the claim that for any $\epsilon >0$ one has that $$\sum_{1 \leq n \leq x}\mu(n) = O(x^{1/2+\epsilon}).$$ This equivalence stems from making an explicit integral for $1/\zeta(s)$ in terms of the Möbius function which converges up the the 1/2 line if one has the above bound on the sum.

Now, it is reasonable to guess/hope/assume that $\mu(n)$ is essentially random in the sense that the non-zero values behave essentially like a fair coin with heads corresponding to 1 and tails corresponding to –1. It turns out that if one has a fair coin, and one keeps flipping it, then the difference between the number of heads and tails after $x$ flips will with probability 1 be $O(x^{1/2+\epsilon})$. So if the Möbius function behaves like a random coin flip, or even close to a random coin flip we should expect RH to hold with probability 1.

Note that this isn't the only thing satisfying about this framing of the Riemann Hypothesis. This sort of shows one major reason that RH is important: a lot of sieves and other ways to get a handle on the primes involve inclusion/exclusion arguments, which is essentially what $\mu(n)$ is for. So in a moral sense RH says that if one is doing inclusion/exclusion with primes, one cannot get very large deviations in how much at any given stage one needs to include or exclude.

(Edit to add: actually this is essentially the same sort of approach as noted in the link by Pace above.)