Now, if my understand is correct, then Van Kampen says $$\pi_1(T) \cong (\mathbb{Z} \ast \mathbb{Z})/N$$
Your understanding is correct, however, the kernel of $\phi$ is not $\langle a,b : aba^{-1}b^{-1}\rangle$. Rather, the kernel is all products of conjugates of $aba^{-1}b^{-1}$, or the "normal closure" of the subgroup generated by $\{aba^{-1}b^{-1}\}$. This is what
when we say "$N$ is generated by the word $aba^{-1}b^{-1}$", we include the conjugation operation
means. Since this may be the point of confusion, I'll expand on it more:
You can think of taking the normal closure of a subgroup as a process: start with the subgroup, and add the necessary elements so that it becomes normal. If $aba^{-1}b^{-1}$ is in the subgroup, and $g$ is some other element of that group, the normal closure must also include $gaba^{-1}b^{-1}g^{-1}$ (ie $aba^{-1}b^{-1}$ conjugated by $g$) if we want it to be normal. The normal closure is basically what you get when you keep on doing this.
Hopefully that clears up what is meant by "including the conjugation operation."
With this view, we can see that $N$ is just all products of conjugates of $aba^{-1}b^{-1}$. (It may take a moment to digest that statement)
Then, when we take the quotient, we can see that $(\mathbb{Z} \ast \mathbb{Z})/N$ is the free group of rank 2 with the added relation $aba^{-1}b^{-1}$. That is,
$$\pi_1(T) \cong (\mathbb{Z} \ast \mathbb{Z})/N \cong \langle a,b : aba^{-1}b^{-1}\rangle \cong \mathbb{Z}^2$$
Edit: in response to your comments: it is not true that $(\mathbb{Z} \ast \mathbb{Z})/N \cong N$.
Second edit: As for actually computing the quotient...
First, think about the coset $aba^{-1}b^{-1}N = aNbNa^{-1}Nb^{-1}N$. This must be equal to $N$, as $aba^{-1}b^{-1} \in N$. Then, $aNbNa^{-1}Nb^{-1}N = N$, so $aNbN = bNaN$, ie the quotient is abelian. So, we add the relation $aNbN = bNaN$. Now, are there any other relations? no:
If $wN = N$, then $w \in N$, so $w$ is a product of conjugates of $aba^{-1}b^{-1}$. Then, $wN$ is a product of conjugates of $aNbNa^{-1}Nb^{-1}N$, which is just the identity (in light of the previous paragraph). Then $aNbN = bNaN$ is the only relation, and we get the presentation $$ \langle a,b : aba^{-1}b^{-1}\rangle$$
Best Answer
Consider a 1-genus torus with two boundary components:
and the paths $a,b,w_1,w_2$ as shown here:
Since you say
I'll assume you are somewhat comfortable with the fact that the boundary of $B$ (the torus with a single hole) can be expressed as $efe^{-1}f^{-1}$. It is a similar fact that $w_1$ and $w_2$ differ by the commutator $[a,b] = aba^{-1}b^{-1}$. That is, $w_1aba^{-1}b^{-1} = w_2$. I encourage you to draw this all out and convince yourself of this. (I may have drawn the orientations backwards, so you may have to throw in some inverses to make it work.)
Now consider $A$:
Here, I've drawn in some additional curves to help us out. Strictly speaking, these should all be attached to some basepoint as they were in the previous picture, but I didn't want to clutter it. Also for the sake of clutter reduction, I've omitted $a,b,c,d$, where $a,b$ are the paths as in the previous picture around the leftmost hole, and $c,d$ are those around the rightmost.
By applying the first paragraph to the leftmost hole, we know that $w_1aba^{-1}b^{-1} = w_2$. Similarly, for the rightmost hole, we have $w_1 = wcdc^{-1}d{-1}$. Substituting, we get $w_2 = wcdc^{-1}d{-1}aba^{-1}b^{-1}$. But $w_2$ is homotopically trivial, so $w = bab^{-1}a^{-1}dcd^{-1}c^{-1}$.
As before, this depends on exactly what the orientations of everything are, so you might have to throw in some inverses. Accounting for that, (I think) this is what you wanted.
Hope it helps.
I'll also mention that the usual (and easier) way of finding the fundamental group of the $n$-genus torus is by using the $4n$-gon identification, an outline of which can be found here.