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.
After filling in the boundary with a handlebody you can basically flatten the handlebody by isotopy to collapse $U_\beta$ and $U_\gamma$. If you think of each of $U_\beta$ and $U_\gamma$ as a thickened 3d handlebody then you push them together to get an even thicker 3d handlebody. So your original cobordism was a Y shape, but you push two of the spokes together to get something with only two spokes, and both "boundary sides" are determined by the same two sets of curves (technically one side by $\alpha$ and $\beta$ curves and the other side by $\alpha$ and $\gamma$ curves but the $\beta$ and $\gamma$ curves are isotopic).