Given a smooth genus 2 curve $C$, it is canonically a two-fold cover of a $\mathbb{P}^1$, branched at six points. Allowing stable curves allows degenerations in two directions: the six points are allowed to collide (in certain ways), and $\mathbb{P}^1$ can "break" into a degenerate conic in $\mathbb{P}^2$ (here a union of two distinct lines). For instance, a smooth genus one curve with two of its points glued together is an example of the former situation, while two genus one curves joined to each other along a point is an example of the latter.
It is possible to construct explicit families (ideally, over high-dimensional bases and without appealing to stable reduction results) of stable genus 2 curves that include degenerations of both kinds? (I would even be interested in degeneration of smooth curves to the second locus of unions of elliptic curves.) One difficulty here is that, if $C$ is a union of elliptic curves, then the map from $C$ to the quotient of $C$ by its "hyperelliptic" involution (multiplication by $-1$ on each elliptic curve) is not flat at the singular point.
Best Answer
As Akhil points out, you can have both types of behavior simultaneously. For instance, consider a curve $C$ that is a union of two irreducible components $C'$ and $C''$ that intersect transversally at a single point $p$ (smooth on each component). Let $C'$ be a smooth, genus $1$ curve. Let $C''$ be a nodal genus $1$ curve with geometric genus $0$. Let $\nu':C'\to B'$ be a double cover of a genus $0$ curve ramified at $p$. Let $\nu'':C''\to B''$ be a double cover of a genus $0$ curve ramified at $p$. Then there is a unique at-worst-nodal curve $B=B'\cup B''$ of genus $0$, glued together at $\nu'(p)=\nu''(p)$. Also there is a unique morphism $\nu:C\to B$ whose restriction to $C'$ is $\nu'$ and whose restriction to $C''$ is $\nu''$.
In some sense, this already shows how to produce explicit families: deform the pointed curve $(B,D)$, where $D$ is the union of the branch points of $\nu$ (not counting $p$). This can be made very explicit: first, identify $B$ with $(B'\times\{p\})\cup (\{p\}\times B'')$ inside $B'\times B''$. Now deform this curve in a general pencil of divisors $(\mathcal{B}_s)_{s\in S}$ in $B'\times B''$. The divisor $D$ is the intersection in $B'\times B''$ of $B$ and a divisor $E$ of type $(3,3)$. So also deform $E$ in a general pencil $(\mathcal{E}_t)_{t\in T}$ of divisors of type $(3,3)$ in $B'\times B''$. Now, over the base $S\times T$, consider the family of curves $\mathcal{B}_s$ with divisors $\mathcal{D}_{s,t} = \mathcal{B}_s \cap \mathcal{E}_t$.
The only issue is that, in order to explicitly form the double cover $\mathcal{C}_{s,t}\to \mathcal{B}_t$ branched over $\mathcal{D}_{s,t}$, we need an invertible sheaf on $\mathcal{B}_t$ and an explicit isomorphism of this invertible sheaf with $\mathcal{O}_{\mathcal{B}_t}(\mathcal{D}_{s,t})$. We can form this ideal sheaf on the generic fiber. However, filling in the branched cover almost certainly will require passing to a degree $2$ cover of $S\times T$ (at least, that is my recollection).
Edit. I can be a little more explicit. For a general pencil $S$ as above, there are precisely two singular members of the pencil, i.e., there are two $k$-rational points, $0,\infty \in S$, such that for $S^* = S\setminus\{0,\infty\}$, the restricted morphism $\pi^*:\mathcal{B}^* \to S^*$ is smooth and projective. Moreover, there are sections: the two base points of the pencil. So $\mathcal{B}^*\to S^*$ is the projectivization of a rank $2$, locally free sheaf on $S^*$. Note that $S^*$ is isomorphic to $\mathbb{G}_m$, and thus has trivial Picard group. Therefore, $\text{Pic}(\mathcal{B}^*)$ is a free Abelian group of rank 1, with an ample generator $\mathcal{A}$ that has degree $1$ on the geometric fibers of $\pi^*$. In particular, the pullback from $B'\times B''$ of $\mathcal{O}(3,3)$ is isomorphic to $\mathcal{A}^{\otimes 6}$. Therefore, over $S^*\times T$, the ample invertible sheaf we are trying to "halve" is $\mathcal{A}^{\otimes 6}\boxtimes \mathcal{O}_T(1).$ Thus, over each of the two standard open affines in $T$, there is a square root of $\mathcal{O}_{\mathcal{B}^*\times T}(\mathcal{D}^*)$. Hence, over each open affine in $T$, there is also the associated double cover of $\nu^*:\mathcal{C}^* \to \mathcal{B}^*\times T$ branched over $\mathcal{D}^*$. Gluing these together gives a branched cover over all of $\mathcal{B}^*\times T$.
The "discriminant divisor" $\Delta$ of $\mathcal{D} \to S\times T$ is the divisor over which the fibers of the Cartier divisors $\mathcal{B} \times T$ and $S \times \mathcal{E}$ in $(S\times T)\times(B'\times B'')$ are tangent (one can prove that $\Delta$ is a Cartier divisor using sheaves of relative differentials and the "norm" of the finite flat map $\mathcal{D}\to S \times T$). If $T$ is a sufficiently general pencil, then $\Delta$ is a smooth Cartier divisor away from finitely many points of $S\times T$, none of which is the special point $(0,0)$ parameterizing $(B,D)$, and $\Delta$ intersects $\{0\}\times T$ transversally at $(0,0)$. Thus, after deleting finitely many points from $S\times T$ (but not the special point), we may assume that $\Delta \cup \{0,\infty\}\times T$ is a simple normal crossings divisor in $S\times T$.
The family of curves, $\mathcal{C}^*\to S^*\times T$ is a family of smooth curves on the complement of $\Delta$. The fibers over geometric points of $\Delta$ are irreducible curves of arithmetic genus that have a single ordinary double point.
The family of stable curves $\mathcal{C}^*$ over $S^*\times T$ does not extend to $S\times T$. However, it does extend to codimension $1$ points after a ramified base change of order $2$. More precisely, let $u:\widetilde{S}\to S$ be the degree $2$, finite flat morphism that is &eeacute;tale over $S^*$ and is ramified at $0$ and $\infty$. Then the pullback of $\mathcal{C}^*$ to $\widetilde{S}^*\times T$ does extend to codimension $1$ points of $\widetilde{S}\times T$.
Moreover, the pullback $\widetilde{\Delta}$ of $\Delta$ to $\widetilde{S}\times T$ is still transverse to $\{0,\infty\}\times T$ at the special point $(0,0)$, and away from all but finitely many points. Thus, deleting those finitely many points, the full discriminant locus, $\widetilde{\Delta} \cup (\{0,\infty\} \times T)$, is a simple normal crossings divisor in $\widetilde{S}\times T$. Thus, now we can apply the main theorem of de Jong - Oort, "Extending families of curves", to conclude that the family of stable curves extends to a family $\mathcal{C}$ over all of $\widetilde{S}\times T$ (less the finitely many points we deleted).
Since $\widetilde{S}\times T$ is $\mathbb{P}^1\times \mathbb{P}^1$, I think this is about as explicit as possible. One thing to note: for the morphism $\widetilde{\nu}:\widetilde{C}\to \widetilde{B}$, the morphism does fail to be flat over the codimension $2$ locus of "nodes" of $\widetilde{B}$ that projects isomorphically to $\{0,\infty\}\times T$ inside $\widetilde{S}\times T$. So the coherent sheaf $\widetilde{\nu}_*\mathcal{O}_{\widetilde{C}}/\mathcal{O}_{\widetilde{B}}$ is not an invertible sheaf: it is a pure, torsion-free sheaf of rank $1$. It should not be that difficult to work out this sheaf. This would make the family $\widetilde{\mathcal{C}}$ even more explicit.