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.
When I was taught the classification theorem, my professor emphasized being able to compute the Euler characteristic, and his favorite examples that he set as exercises were Seifert surfaces.
For example, you can try to compute the Euler characteristic of the Hopf link: http://en.wikipedia.org/wiki/File:Hopf_band_wikipedia.png.
Or something weirder, like this surface: http://en.wikipedia.org/wiki/File:Borromean_Seifert_surface.png
Both of these images are found in the Wikipedia article for Seifert surfaces. You could easily draw more complicated Seifert surfaces, and their Euler characteristics are not so obvious to compute. The easiest way to compute the Euler characteristic of these two surfaces is to use the inclusion-exclusion principle for Euler characteristic:
$$ \chi(A\cup B) = \chi(A) + \chi(B) - \chi(A\cap B).$$
This lets you decompose the surface into more manageable objects, like disks, lines, and points, for which you should have the Euler characteristic already. So Euler characteristic becomes a recursive calculation.
You can determine orientability of these surfaces fairly easily. Consider a planar projection of the Hopf link surface: it is just the Hopf link, with the region consisting of the surface marked (perhaps shaded). Imagine you color the regions of the Hopf link with black and white, like so: first color any region black. Move to a region neighboring the first region, i.e. separated by a crossing of the link (a twist), and label that region white. Then repeat, switching color each time you move across a twist. If this coloring is consistent, then the surface is orientable. Otherwise, it is non-orientable. So the Hopf link is easily shown by this algorithm to be orientable; on the other hand, the Seifert surface of the trefoil knot is not orientable.
Finally, the number of boundary components is trivial to count.
Once you have all these ingredients, you can use the classification of surfaces with boundary to see what the surface is. I encourage using the Seifert surface of a trefoil to highlight the non-orientable case.
Another way to use the classification theorem is for polygonal representations of surfaces, for example http://en.wikipedia.org/wiki/File:KleinBottleAsSquare.svg. If you have a representation of a surface as a disc with certain pairs of sides identified, then you can easily find its Euler characteristic via the formula $$\chi = V-E+F.$$
Here we need to determine $V$, $E$, and $F$. $F=1$ if we're only considering one disc, since it's the only $2$-dimensional component. For $E$, go around the boundary of the disc and count the number of edges, with identified edges counting once. For $V$, do the same, keeping track of which vertices end up identified. For the Klein bottle example in the picture, we have $F = 1$, $E = 2$, and $V = 1$, so the Euler characteristic is $1-2+1 = 0$. The Klein bottle has no boundary, so the classification theorem states $0 = 2-2g$, so $g=1$. Hence the Klein bottle is closed and nonorientable with genus $1$, which determines its homeomorphism class.
Strictly speaking, this is sort of a special case of surfaces with boundary: the number of boundary components is the number of nonidentified edges. But in any case, this gives you an easy way to generate complicated surfaces with easily computable classifications: take some $n$-sided polygon with $n$ as large as you want, arbitrarily put orientations on the edges, identify pairs of edges, and get to work computing $V$, $E$, and $F$. The surface is non-orientable if there is a pair of identified edges both oriented clockwise or counterclockwise around the disc: basically, if edges can be oriented $+$ (clockwise) and $-$ (counterclockwise), then a non-orientable pair is $++$ or $--$. So orientability is a simple check as well. Find the number of boundary components, and now you have the Euler characteristic, number of boundary components ($0$ if you just make sure all edges are identified with another; you'll need an even-sided polygon to do this), and the orientability of the surface, everything you need to compute the genus. Hence you can classify all such surfaces.
Best Answer
This loop bounds a subsurface (torus with one hole) of genus 2 surface. We know that non-trivial elements of first homology group is 'cycles that are not boundary'. On the other hand since this loop does not bound a disc it is non-trivial on fundemental group.It can be also checked algebraically by drawing planar diagram of genus two surface and see that which element the loop correspond to in fundemental group.