Let me give an answer beynd the case when $X$ is affine.

**Upshot**: Minimally restricting the setup, the answer is "LX is a sheaf, so you don't need to sheafify" for all quasi-projective schemes $X/K((t))$, and with respect to a very strong topology.

Suppose $K$ is has positive characteristic, and restrict your functor LX to the category ${\rm Perf}_K$ of perfect $K$-algebras. Recall the arc-topology on schemes, introduced by Bhatt--Morrow arXiv:1807.04725. Essentially, a map $X \rightarrow Y$ of qcqs schemes is an arc-cover if it's surjective and any specialization relation between points in $Y$ lifts to $X$. E.g., you could take the spectrum of the product over all possible valuation rings at all points of $Y$, thus obtaining a (huge) affine scheme, which maps to $Y$, and this would give you an arc-cover. Note that the arc-topology is even stronger than the $v$-topology of Bhatt--Scholze arXiv:1507.06490, and "much" stronger than the fpqc-topology. E.g. if $V$ is a valuation ring of rank $2$ and $\mathfrak{p}$ is its unique prime ideal of height $1$, then ${\rm Spec} V_{\mathfrak{p}} \,\dot\cup\, {\rm Spec} V/\mathfrak{p} \rightarrow {\rm Spec} V$ is an arc-cover, but not a $v$-cover (and so in particular, not an fpqc-cover), see Corollary 2.9 of arXiv:1807.04725. This said, we have the following result, see Theorem 5.1 of arXiv:2003.04399:

**Theorem.** Suppose $X$ is a quasi-projective $K((t))$-scheme. Then LX is a sheaf for the arc-topology on ${\rm Perf}_K$.

*Remark 1.* I would guess that this should also extend to the case that $K$ has characteristic zero, but the proof in loc.cit. uses perfectoid spaces, and so cannot literally be carried over to the case when $K$ has characterstic zero.

*Remark 2.* This result continues to hold in a mixed characteristic setup, i.e., when $K((t))$ is replaced by an appropriate $p$-adic field with residue field $K$ (e.g. if $K = \mathbb{F}_p$, instead of $K((t))$ you could take $\mathbb{Q}_p$, or some finite totally ramified extension of it).