You can make more examples using the natural $K$-equivalent between the Hilbert scheme of a $K3$ Deligne-Mumford stack to the corresponding Hilbert scheme of the crepant resolution of its coarse moduli space. The example in your post arises this way when the stack has a single stacky point whose inertia group is cyclic of order $2$, so that the crepant resolution has a $(-2)$-curve whose $\text{Sym}^2$ is the flopped $\mathbb{P}^2$ (I assume that your example arises by flopping $\text{Sym}^2$ of one of the preimages of a bitangent line, which is the $(-2)$-curve in my interpretation). The following example is the next case, where the stack has a single stacky point whose inertia group is cyclic of order $3$, so that the crepant resolution is a chain of two $(-2)$-curves.
Let $S$ be a smooth quartic K3 surface in $\mathbb{P}^3$ that is a general member of the $5$-dimensional projective linear system of quartics that contain a specified line $L$ as well as a specified disjoint curve $C=M\cup N$ that is a union of two lines intersection transversally at one point $p$. The plane cubics residual to $L$ form an elliptic fibration on $S$, $$\pi:S\to \mathbb{P}^1.$$ This induces an Abelian fibration on $S^{[2]}$. Let $D'$ denote the pullback to $S^{[2]}$ of the hyperplane class on $\text{Sym}^2(\mathbb{P}^1) = \mathbb{P}^2$.
The curve $M\cup N$ can be contracted to form a normal projective surface $S'$ with a single $A_2$-singularity $q$. Explicitly, the $2$-plane spanned by $M$ and $N$ intersects $S$ in the union of $M$, of $N$, and of a plane conic $C$. The linear system of sections of $\mathcal{O}_{\mathbb{P}^3}(2)$ that contain $C$ induces a closed immersion of $S'$ as a sextic surface in $\mathbb{P}^4$ (probably there is a direct construction of this sextic suface).
There is a smooth, proper Deligne-Mumford stack $\mathcal{S}$ whose coarse moduli space is a $1$-morphism, $$\nu:\mathcal{S}\to S',$$ that is an isomorphism over the open subset $U:=S'\setminus\{q\}$, and whose inertia group over $q$ is cyclic of order $3$. The Hilbert scheme $X$ of length $2$ closed substacks of $\mathcal{S}$ has a unique connected component that contains $U^{[2]}$. This is a projective hyperkähler $4$-fold that is K-equivalent, hence deformation equivalent (by Huybrechts), to $S^{[2]}$.
The surface in $X$ that is the fundamental locus of the K-equivalence with $S^{[2]}$ equals the punctual Hilbert scheme of $\mathcal{S}$ for the point $q$. Because $K$-equivalences preserve classes in the Grothendieck ring of varieties, etc., we know that this fundamental locus has the same class as the fiber over $2[q]$ for the Hilbert-Chow morphism, $$FC:S^{[2]}\to \text{Sym}^2(S').$$ In particular, the fundamental locus has $3$ irreducible components, each of which is a rational surface.