Let $X$ be a scheme over $K = k((t))$, where $k$ is a field. We define the loop scheme $LX$ to be the functor from the category of $k$-algebras to sets by $R \mapsto LX(R) := X(Spec (R((t))))$.
Do we need sheafification in order to make $LX$ a sheaf? A particular interest is when $X$ is a connected linear reductive group over $K$ and fpqc site is chosen.
Despite Will Sawin's answer in comment, it was not so obvious for me at first glance, so I give a detailed (at least for $X$ affine) proof here.
Lemma. Let $A$ be a $K$-algebra and let $A\to B$ be fpqc. Then the following sequence is exact $$ A((t))\to B((t))\rightrightarrows (B\otimes_{A}B)((t)). $$
Proof. Since $A\to B$ is fpqc, Grothendieck's fpqc-descent shows that the sequence $$ A\to B\rightrightarrows B\otimes_A B $$ is exact. Consider the sequence $$ A[[t]]\to B[[t]]\rightrightarrows (B\otimes_{A}B)[[t]]. $$ We claim that this sequence is also exact.
The first arrow $A[[t]]\to B[[t]]$ is injective since $A\to B$ is injective and since $A[[t]] = \prod_{n\in \mathbf{N}}A$ and $B[[t]] = \prod_{n\in \mathbf{N}}B$ as $A$-modules. It remains to show that for $f(t)\in B[[t]]$, $$f(t)\otimes 1 = 1\otimes f(t)\in (B\otimes_{A}B)[[t]] \quad\Longrightarrow \quad f(t)\in A[[t]]. $$ Suppose that $f(t) = \sum_{n\in \mathbf{N}} b_n t^n\in B[[t]]$ satisfies the condition $f(t)\otimes 1 = 1\otimes f(t)$. It is enough to show that $f(t) \mod{t^N} \in A[t]/t^N$ for each $N\ge 1$. Indeed, if we tensorise the first exact sequence by $\otimes_{A}A[t]/t^N$, we get $$ A[t]/t^N\to B[t]/t^N\rightrightarrows (B\otimes_{A} B)[t]/t^N, $$ which implies that $b_n\in A$ for $n < N$. Hence $f(t)\in A[[t]]$.
Inverting $t$, we deduce the exactness of $$ A((t))\to B((t))\rightrightarrows (B\otimes_{A}B)((t)). $$ End of Proof
Now if $X = \mathrm{Spec}\, C$ is an affine $K$-scheme and A,B are as above, then the exactness of the following sequence follows from the lemma $$ \mathrm{Hom}(C, A((t)))\to \mathrm{Hom}(C, B((t)))\rightrightarrows \mathrm{Hom}(C, (B\otimes_{A} B)((t))). $$ Hence, if $X$ is affine, $LX$ is automatically an fpqc sheaf. The general case follows from the standard procedure of glueing.
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).
