$\newcommand{\co}{\mathcal{O}} \newcommand{\H}{\mathrm{H}} \newcommand{\dR}{\mathrm{dR}} \newcommand{\GG}{\mathbf{G}}$
Let $k = \mathbf{C}$, let $X$ be a (smooth) variety over $k$, and let $X_\dR$ be its de Rham space (so that the cohomology of $\Gamma(X_\dR; \co_{X_\dR})$ is $\H^\ast_\dR(X/k)$. There is a canonical map $f: X \to X_\dR$, and the global sections of the fiber of the map $\co_{X_\dR} \to f_\ast \co_X$ is $\H^\ast(X; \Omega^{\bullet\geq 1}_{X/k})$. In other words, the map $\co_{X_\dR} \to f_\ast \co_X$ corresponds to the projection $\Omega^{\bullet}_{X/k} \to \co_X$ (whose fiber is $\Omega^{\bullet\geq 1}_{X/k}$, with $\Omega^i_{X/k}$ placed in homological degree $-i$). From this point of view (some of the indices may be off, let me know if you find mistakes):

The group $\H^2(X; \Omega^{\bullet\geq 1}_{X/k})$ may be understood as $\pi_0$ of the space of $\GG_a$-gerbes on $X_\dR$ which pull back to the trivial $\GG_a$-gerbe along $f$. Note that $\H^2(X; \Omega^{\bullet\geq 1}_{X/k}) = \H^1(X; K_1)$ in your notation. (The relation to TDOs comes from an identification of $\GG_a$-gerbes on $X_\dR$ which pull back to the trivial $\GG_a$-gerbe along $f$ with $\GG_m$-gerbes on $X_\dR$ which pull back to the trivial $\GG_m$-gerbe along $f$, using the exponential on the formal completion of $\GG_a$. Because $\mathrm{QCoh}(X_\dR)$ describes D-modules on $X$ (using the finite-type assumption on $X$), $\GG_m$-gerbes on $X_\dR$ correspond to twistings of $\mathrm{DMod}(X)$. See Gaitsgory-Rozenblyum's https://arxiv.org/abs/1111.2087 for more on this.)

Similarly, $\H^3(X; \Omega^{\bullet\geq 2}_{X/k}) = \H^1(X; K_2)$ in your notation, and this can be understood as $\pi_0$ of the space of $\GG_a$-$2$-gerbes on $X_\dR$ which pull back to the trivial $\GG_a$-$2$-gerbe along $f$.

In general (if I haven't snuck in a $\pm 1$ mistake), if we define the space of $\GG_a$-$n$-gerbes on $Y$ to be the space of maps $Y \to B^{n+1} \GG_a$, then $\H^{n+1}(X; \Omega^{\bullet\geq n}_{X/k}) = \H^1(X; K_n)$ in your notation, and this can be understood as $\pi_0$ of the space of $\GG_a$-$n$-gerbes on $X_\dR$ which pull back to the trivial $\GG_a$-$n$-gerbe along $f$.

For this to be a satisfying answer, we should relate $\GG_a$-$2$-gerbes on $X_\dR$ to chiral differential operators . This should be given by an algebraic version of transgression (which, in topology, would be a map $\H^\ast(Y) \to \H^{\ast-1}(LY)$ where $Y$ is an oriented manifold). I don't know a reference for this in the algebraic setting (but I would like to know of one if it exists!), but if you are interested I can try to sketch the idea. I think $K_n$ wouldn't be related to iterated free loop spaces, but rather to some algebraic analogue of $\mathrm{Map}(S^n, Y)$.

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