Let's call a double of a 2-handlebody a *2-double* for simplicity. I will tacitly assume orientability of everything.

Your first question is interesting, I only have some partial remarks. 
As you noticed if $X^4$ is a double then it must be null-cobordant hence $\sigma(X) = 0$, and we can also add that the Euler characteristic $\chi(X) \in 2\mathbb Z$ has to be even.

At this point we can already conclude that: **the only simply connected rational surfaces which are 2-doubles are $\mathbb S^2\times \mathbb S^2$ and $n\mathbb {CP}^2\#n \overline{\mathbb{CP}}^2$**. Indeed  the constraint on $\sigma$ and $\chi$ imply that these are the only possibilities and they clearly are 2-doubles since $\mathbb S^2\times \mathbb S^2$ and $\mathbb {CP}^2\# \overline{\mathbb{CP}}^2$ are the doubles of the two $\mathbb D^2$-bundles over $\mathbb S^2$ and connected sums of two 2-doubles is a 2-double. 

If we consider the case $b^+\geq 2$ then we have another constraint from Seiberg-Witten theory: by Kirby calculus it is not difficult to see that if a 2-double has $b_2(X)>0$ then exist an embedded 2-sphere $S\subset X$ such that $S^2 = 0$ and $[S]\neq 0 \in H_2(X)$. To see this, let $X = D(Z)$, $Z$ a 2-handlebody and consider a Kirby diagram for $Z$. Since $H_2(X)\neq 0$,  up to handle-slides we can assume that exists a 2-handle of $Z$, $h$ representing a non-trivial homology class. To construct the double we glue 2-handles along 0-framed meridians of each 2-handle attaching circle (and then we add some 3-handles). The 0-framed meridian associated to $h$ will induce a non-trivial homology class in $H_2(X)$ (represented by a sphere) because it cannot be in the image of the three-handles for otherwise $h$ would not be a 2-cycle of $C_2(Z;\mathbb Z)$. 

Now it follows from Fintushel & Stern. *Immersed spheres in 4-manifolds and the immersed Thom conjecture*, Turkish J. Math., 19(1995), no. 2, 145–157 that
**if $X$ is a 2-double and  $b^+(X)\geq 2$ then $X$ has vanishing Seiberg-Witten invariants**.
Consequently, by the work of Taubes we obtain that **simplectic 4-manifolds and complex surfaces with $b^+\geq 2$ cannot be 2-doubles**.



Regarding your last question we can show the following: **for any simply connected $X^4$ exists a simply connected $N$ such that $X\#N$ is a 2-double**.

*Proof*:  by blowing up and down $X$ we can make its intersection form isomorphic to the intersection form of  $n\mathbb {CP}^2\#n \overline{\mathbb{CP}}^2$, thus by Freedman's theorem is also homeomorphic to it. Moreover by Wall's theorem we can  stabilize with a certain number of $\mathbb S^2\times \mathbb S^2$ and obtain diffeomorphic manifolds. But the result of this stabilization is a 2-double for the first proposition above.