You can get the genus $g$-surface by doing the connected sum of $g$ tori $T=S^1 \times S^1$, i.e.,
$$S_g := T\,\#\,T\,\#\,\cdots\,\#\,T \qquad (g\text{ times}).$$
Assuming you're working over $ \mathbb{Z}$.
If you know the homology of $T$, and how to find that of the connected sum, done.
If not, or if you prefer a different approach, you can:
i) Find the homology of $S^1$, then
ii) Find the homology of the product $S^1\times S^1$, and
iii) Find the homology of the connected sum of $g$ copies in ii):
$H_1(S^1) = \mathbb{Z}$
How to find the homology of a product space, (e.g., Künneth's formula)
it is $\mathbb{Z}^2$
Finding the homology of connected sums; it is the direct sum of the respective
homologies; the basis curves are pairwise disjoint, so the homology is the
direct sum (what happens in one Torus, stays in that Torus)
You ultimately get: $$H_1(S_g)=\mathbb{Z}^{2g}$$ you are done.
The short answer is your generators are correct. I'm not quite sure how to draw images on this site yet (my first post!), so I'll try to explain in just words. Perhaps you could draw your own pictures. My apologies if this is long-winded, but I am trying to provide you with your sought-after practical description.
First, let's think about a genus $g$-surface, which we imagine as a $2g$-gon with edges cyclically labelled $a_1, b_1, a_1^{-1}, b_1^{-1},\dots, a_g, b_g, a_g^{-1}, b_g^{-1}$. All this means is that we can use this labeling to glue the edges of the $2g$-gon to get our genus $g$-surface. I'll assume that you know how to get the homology of this closed surface, and know that it is the rank $2g$ free abelian group generated by $a_1,b_1,\dots,a_g,b_g$.
So what happens to this picture if we add a single boundary component? To do this, we can remove a small open disk $D_1$ from the interior of our labelled $2g$-gon to get a surface $\Sigma_{g,1},$ where $\partial\Sigma_{g,1}=\partial D_1=:B_1\simeq S^1$. As it turns out and as you are well aware, in this case for $b=1$, the first homology doesn't change; that is, $H_1(\Sigma_{g,0})=H_1(\Sigma_{g,1})$, and their generators are the same. Let's see why.
Note that our $2g$-gon with a disc removed deformation retracts to the labeled edges of the polygon. Since nothing weird is going on at the edges (i.e. the deformation retract is the identity here), we can push this deformation retract forward to the glued quotient space. Thus, we get a deformation retract $r$ from $\Sigma_{g,1}$ to a wedge of $2g$ circles $\bigvee_{i=1}^{2g}S^1$. Since a deformation retract is a homotopy equivalence we have another map $$i:\bigvee_{i=1}^{2g}S^1\rightarrow\Sigma_{g,1}$$
such that $$i\circ r\simeq\mathbb{1}_{\Sigma_{g,1}} \text{and} r\circ i\simeq\mathbb{1}_{\bigvee_{i=1}^{2g}S^1}.$$
In fact, if we identify $\bigvee_{i=1}^{2g}S^1$ with the image of our surface after the deformation retract, we can take $i$ to be the inclusion map, in which case we know that the generators of $H_1(\bigvee_{i=1}^{2g}S^1)$ are precisely the generators of $H_1(\Sigma_{g,1}),$ namely $a_1,\dots,b_g.$
Trying to generalize this deformation retract approach to $b>1$ is a tad difficult, so I'll explain it intuitively first. Start with our labelled $2g$-gon and draw $b$ circles inside. Start blowing up one of the circles like a balloon, without letting it cross itself. We see that the farthest the balloon can go is to completely cover the edges of the polygon and the $b-1$ other boundary components, in the process introducing some new edges between the boundary components and the edges of the polygon. Of course, as before, nothing weird is going on at the edges of the polygon, so we're really doing all of this on the glued-up surface. The resulting image of the balloon--or (dropping the analogy) the image of our surface after the deformation retract--is now a graph with edges given by $a_1,b_1,\dots,a_g,b_g$ and the $b-1$ boundary components we didn't move, as well as the new edges we introduced in blowing up our balloon. However, with some imagination we can convince ourselves that the new edges we introduced add no new cycles because we didn't allow our balloon to pass through itself. Thus we can deformation retract all of the new edges we introduced to get a wedge of $2g+b-1$ circles.
This can all be made rigorous by showing that we can consider our surface with boundary as being obtained by gluing a disk with a hole in it to a graph with $2g+b-1$, essentially working backwards from the intuition we built from the previous paragraph.
For the sake of a complete picture (without actually providing one!) here's how I imagine doing this. Start, as always, with the labelled $2g$-gon and draw $b-1$ circles $c_1,\dots,c_{b-1}$ inside of it, lining them up nicely from left to right in a nicely ordered fashion. Now draw an edge from $c_i$ to $c_{i+1}$ for $1\leq i\leq b-2$, making sure not to cross any of the edges (that would just be silly!). Draw one final edge from $c_{b-1}$ to the nearest vertex of the polygon, again making sure not to add any crossings. We now see before us a connected graph, whose complement in the pre-gluing polygon is a disk. After gluing according to the edge labeling, we get a surface with boundary with a graph drawn on it such that the complement of the graph is a disk. Thus, if we introduce a $b^{\rm th}$ boundary component, we have to remove a smaller disk from the interior of this disk, allowing us to construct a deformation retract from our surface with $b$ boundary components to the graph we drew before (i.e. the graph that tells us how to glue in the punctured disk). We can now use the same argument for the $b=1$ case to show that your generators are correct.
The lesson is that while we can generate $H_1$ of our surface by all of the $2g$ latitude/longitude generators $a_1,b_1,\dots,a_g,b_g$ and all of the $b$ boundary curves $B_1,\dots,B_b$, the deformation retract shows that (assuming appropriate orientations or working over $\mathbb{Z}/2\mathbb{Z}$) we can always identify, say $B_b$ with a sum of all of the other boundary components.
Best Answer
Let $D(E) \to B$ be the unit disc bundle of a vector bundle $E$ (with respect to some metric), and denote the image of the zero section $\sigma$ by $Z$.
The continuous map $H : D(E)\times[0,1] \to D(E)$ given by $(e, t) \mapsto (1-t)e$ satisfies
Therefore $H$ is a strong deformation retraction of $D(E)$ onto $Z$. Note that $\pi|_Z : Z \to B$ is a homeomorphism (with inverse given by $\sigma$), so $D(E)$ is homotopy equivalent to $B$.
In your case, it follows that $H_*(X) = H_*(\Sigma_g\setminus D^2)$. As $\Sigma_g\setminus D^2$ deformation retracts onto a bouquet of $2g$ circles, we see that $H_0(X) \cong \mathbb{Z}$, $H_1(X) \cong \mathbb{Z}^{2g}$, and $H_k(X) = 0$ for all other $k$.
Beyond the homotopy type, note that the normal bundle of $\Sigma_g\setminus D^2$ in $\mathbb{CP}^2\setminus D^4$ is an orientable rank two vector bundle over $\Sigma_g\setminus D^2$. As such, it is determined up to isomorphism by its Euler class in $H^2(\Sigma_g\setminus D^2) = 0$. So the normal bundle, and the corresponding disc bundle, is trivial. Therefore $X$ is diffeomorphic to $D^2\times(\Sigma_g\setminus D^2)$.