I'm not sure I can give you a morally satisfying answer. To my mind, this sort of theorem should be true because it works over $mathbb{C}$, for topological reasons. Of course, that isn't a proof over other fields. But, to my limited understanding, the intuition comes from $mathbb{C}$ and the proofs are motivated by taking proofs which work over $mathbb{C}$ and seeing whether we can generalize them to an arbitrary field.
That said, I can sketch two proofs of this theorem. The first proof is by induction on dimension. Write $X$ as a family over $Y$, with fibers of dimension $dim X - dim Y$. A spectral sequence shows that, if the cohomology of $Y$ vanishes is in degree $>i$ and the cohomology of the fibers vanishes in dimension $>j$, then the cohomology of $X$ vanishes in degree $>i+j$. This reduces us to showing that curves have no cohomology above $H^2$, which can be done by hand; I think this is in Chapter 14 of Milne's lectures.
The other method is for deRham-like cohomology theories (rigid, crystalline, etc.). Roughly speaking, those methods compute cohomology as the hypercohomology of a complex $Omega^0 to Omega^1 to Omega^2 to cdots$. If your ground field is characteristic zero, and your variety is smooth, these are actually the familiar sheaves of differentials. If one of these two conditions fails, you have to adjust in some manner; the details of this adjustment describe which of the theories you are working in. In any case (very roughly speaking) the $2n$ appears here as $n+n$: the complex has length $n$ and each of the terms in the complex has no cohomology above degree $n$. Again, a spectral sequence finishes the proof from here.
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