I'll take the easiest question first: The sum is a so-called formal sum. That means basically that the addition means as little as it can. For instance, $[l]+[m]$ means "We have one copy of the class $[m]$, and one copy of the class $[l]$." The sum $3[m] - 2[l]$ means "We have two copies of the class $[m]$ and negative three copies of the class $[l]$."
As for the generators, here I have to go a bit in-depth about what homology groups are. They are a quotient group of a certain subgroup of the group $C_n$ of $n$-chains. Exactly what these $n$-chains are depends on which interpretation of homology you're using, but in singular homology, they're (formal) sums of functions $f:\Delta^n\to X$, where $X$ is your space, $\Delta^n$ is the standard $n$-dimensional simplex.
We also have a so-called boundary map $d$ that takes an $n$-chain to an $n-1$-chain by restricting any $f$ to all the (oriented hyper-)faces of $\Delta^n$ in turn (with alternating signs). For instance, if $f:\Delta^2\to X$ is a triangle, then $d(f)$ is the "alternating sum of its edges, with orientation". The important thing about which signs and which orientations you use is this: Given any $n$-chain $c = a_1f_1 + \cdots + a_kf_k$, we must have $d(d(c)) = 0$ as an $(n-2)$-chain.
The subgroup of $C_n$ given by all $n$-chains $c$ for which $d(c) = 0$ is called the group of cycles, $Z_n$. A typical example of an element in $Z_1$ is a loop, in $Z_2$ it's a sphere, and so on. The subgroup of $C_n$ consisting of all chains $c$ for which there is an $(n+1)$-chain $c'$ with $d(c') = c$ is called the group of boundaries. A typical example is the boundary of a disc for $n = 1$, or the boundary of a ball, for $n = 2$. The above "$d(d(c)) = 0$ for all $c$" means that we have $B_n \subseteq Z_n$. The $n$-th homology group, $H_n$, is defined as the quotient $Z_n/B_n$. It represents the cycles which cannot be "filled in" within your space.
The generators of the homology group are usually taken to be an equivalence classes represented by a specific cycle-which-is-not-a-boundary. In your example, we choose the representatives $m$ and $l$, since these are two loops that are not the boundary of a disc. Of course, as with any group, you can't choose any set as generators. You need to choose elements that actually generate the group. But that's more of a question in general group theory, not homology.