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.

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 $\nu'(b)=\nu''(b)$. 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).

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).