In the paper by Ozsvath and Szabo, they consider a 4-manifold with boundary which is constructed from a Heegaard triple, i.e., a closed Riemann surface $\Sigma$ with three collections of simple closed curves $\alpha, \beta, \gamma$, each of which generates a half-dimensional sublattice of $H_1(\Sigma)$ (so that each of $\alpha, \beta,\gamma$ determine a handlebody with boundary $\Sigma$ by collapsing the curves). Let's denote this data as $(\Sigma, \alpha,\beta,\gamma)$. Furthermore let's denote the handlebodies obtained from $\alpha,\beta,\gamma$ as $U_\alpha,U_\beta,U_\gamma$. The construction is as follows:
Let $\Delta$ be a (topological) triangle, and let the three sides of $\Delta$ be $e_\alpha,e_\beta,e_\gamma$, resp. Then we can consider a space
$X_{\alpha,\beta,\gamma}=\frac{(\Delta\times \Sigma )\sqcup (e_\alpha\times U_\alpha)\sqcup (e_\beta\times U_\beta)\sqcup (e_\gamma\times U_\gamma) }{e_\alpha\times \Sigma \sim e_\alpha\times \partial U_\alpha, e_\beta\times \Sigma \sim e_\beta\times \partial U_\beta, e_\gamma\times \Sigma \sim e_\gamma\times \partial U_\gamma}$, where each of the gluing $e_i\times \Sigma \sim e_i \times \partial U_i$ is identifying $\partial U_i$ with $\Sigma$ in obvious way.
Then it is obvious that $X_{\alpha,\beta,\gamma}$ is a 4-manifold with boundary $Y_{\alpha,\beta},Y_{\beta,\gamma},Y_{\gamma,\alpha}$ where $Y_{\alpha,\beta}=U_\alpha\cup U_\beta$ is the 3-manifold obtained from the Heegaard diagram $(\Sigma,\alpha,\beta)$, and so on. (In terms of the natural projection $X_{\alpha,\beta,\gamma}\to \Delta$, the boundaries are the preimage of 3 vertices of $\Delta$.)
In the above, I was careless on the orientations, but it is also easy to see that with the natural choice of orientations on the 3-manifold obtained from a Heegaard diagram and the natural choice of the orientation of product manifolds, $\partial X_{\alpha,\beta,\gamma}=-Y_{\alpha,\beta}-Y_{\beta,\gamma}+Y_{\alpha, \gamma}$, i.e., $X_{\alpha,\beta,\gamma}$ can be thought of as a cobordism from $Y_{\alpha,\beta}\sqcup Y_{\beta,\gamma}$ to $Y_{\alpha,\gamma}$.
Then Example 8.1. claims that if we choose $\gamma$ to be an isotopic copy of $\beta$, then the cobordism $X_{\alpha,\beta,\gamma}$ is obtained from the identity cobordism $Y\times [0,1]$ by deleting a regular neighborhood of $U_\beta \times \left\{\frac{1}{2}\right\}$, or equivalently a regular neighborhood of a bouquet of $g$ circles. (where $Y=Y_{\alpha,\beta}$)
Q. How can you visualize this example? I mean, of course $Y_{\beta,\gamma}=\#^g (S^1\times S^2)$ in this case, which bounds a boundary connected sum of $g$ copies of $S^1\times B^3$, but it is not obvious to me that $X_{\alpha,\beta,\gamma}$ becomes $Y\times [0,1]$ after filling out the $Y_{\beta,\gamma}$ boundary by it.
