What is the line bundle $\mathcal{O}_{X}(k)$ intuitively?

Here is a rather category-theoretic explanation where these Serre twists come from. When the graded ring $A$ is finitely generated by $A_1$ over $A_0$, there is a well-known universal property of $\mathrm{Proj}(A)$. Namely, morphisms $Y \to \mathrm{Proj}(A)$ over $A_0$ correspond bijectively to line bundles $\mathcal{L}$ on $Y$ together with an $A_0$-linear epimorphism $A_1 \to \Gamma(Y,\mathcal{L})$. This describes the functor of points $\hom(-,\mathrm{Proj}(A))$. Therefore, actually this may serve as a definition of the Proj construction. Now the universal element of this representable functor is a line bundle $\mathcal{O}(1)$ on $\mathrm{Proj}(A)$ together with a epimorphism $A_1 \to \Gamma(\mathrm{Proj}(A),\mathcal{O}(1))$. More generally, $\mathcal{O}(k) := \mathcal{O}(1)^{\otimes k}$ for $k \in \mathbb{Z}$.

Intuitively, Serre twists make it possible to "shift to the affine case". If $\mathcal{F}$ is a coherent sheaf on an affine scheme, there is an epimorphism $\mathcal{O}^n \twoheadrightarrow \mathcal{F}$ (global generators). This is not true in the projective case. However, for every coherent sheaf $\mathcal{F}$ on a projective scheme $X$ with a choosen ample sheaf $\mathcal{O}(1)$ there is an epimorphism $\mathcal{O}^n \twoheadrightarrow \mathcal{F}(k) := \mathcal{F} \otimes \mathcal{O}(k)$ for $k$ large enough. Intuitively, we are just clearing denominators here in order to reduce to the affine situation. It follows that there is an exact sequence

$\mathcal{O}(-k_2)^{n_2} \to \mathcal{O}(-k_1)^{n_1} \to \mathcal{F} \to 0$

In the affine case, we have $k_1=k_2=0$ and this would be a description by generators and relations. Here, this is something really similar, we just have added degrees to the generators. This also shows that the category of coherent sheaves is generated by the Serre twists.

Similarily, cohomology vanishes after shifting high enough, etc.