Let M be a 3-dimensional homotopy sphere. Remove two embedded 3-disks. One obtains a simply connected $h$-cobordism $W$ from S^2 to S^2. By Morse theory it admits a handle body decomposition. Using transversality one can order the handles in ascending indexes. Further more one can always clear 0-handles canceling with 1-handles. Using the dual decomposition one can also clear 3 handles. Now the number $g$ of 1- and 2- handles has to agree. Hence
$$W \cong S^2 \times [0,1] + \sum\limits_{k=1}^{p_1} (\phi_k^1) + \sum\limits_{k=1}^{p_2} (\phi_k^2)$$
Gluing back one $D^3$ one obtains a configuration where to a handle body of genus $g$ are $g$ 2-handles are attached along simple disjoint curves. Fix a base point in the boundary and choose paths from that point to points in the image of each curve. These curves have to form a basis of the fundamental group of the handle body in order to render the final manifold simply connected.
According to this paper attaching a 2-handle to a handle body along one of these yields a handle body of genus one smaller. Thus after attaching all $g$ 2-handles one obtains a handle body with no 1-handles, ie. a $D^3$. Finally $M$ is obtained by gluing in the remaining $D^3$ back along a homeomorphism of $S^2$.
I know I have to be missing something!
Here is what you should be aware of. Let $x_1,...,x_n$ be a free basis of $F_n$. Then there exist elements $y_1,...,y_n$ such that $$ x_1'=y_1 x_1 y_1^{-1},..., x_n'=y_n x_n y_n^{-1} $$ is not a basis of $F_n$. (The elements $x_1',...,x_n'$ generate a proper subgroup of $F_n$.) The core curves along 2-handles are attached define elements of $F_n$ only up to conjugation (choices of approach paths from the base-point).