In standard topological terms, the exact sequence that relates homotopy groups of the base $B$, fiber $F$ and total space $E$ of topological fibration gives
$$\pi_1(F) \to \pi_1(E) \to \pi_1(B) \to \pi_0(F),$$
that is,
$$ 0\to \pi_1(T-4) \to \pi_1(S^2-4) \to \mathbb Z_2.$$
The middle map works by taking a loop above and pushing it to the base; the right map works by taking a loop on $S^2-4$, lifting it as a path, and taking 0 or 1 depending on whether the resulting path is closed or not.
I believe the OP described the fundamental group of the torus as $\left< a, b, c, d, e, f| [a, b]cdef = 1\right>$ where $a, b$ are two circles of the torus and $c, d, e, f$ are four loops around the holes. For $S^2-4$ my suggestion would be to use $\left< C, D, E, F| CDEF = 1\right> = \left< C, D, E\right>$ where $c$ is over $C$ etc (rather then $g, h, x, w$).
Now, since the loop around $c$ on torus has to wind twice around it when projected to the sphere (think about complex $z\mapsto z^2$ map) it's easy to see that there should be something like $c\mapsto C^2$, $d\mapsto D^2$ etc.
What about $a$ and $b$? Careful observer should note that we haven't defined the loops well yet: if you move $a$ parallel to itself, you'll get a new $a'$ which would differ by something like $cd$ depending on which points are where and depending on how you draw the basepoints on the loops.
For the exact calculations one should fix the torus to be $\mathbb R\times \mathbb R/\mathbb Z\times\mathbb Z$ so that fixed points of $z\mapsto z$ are the vertices $c = (0, 0)$, $d = (1/2, 0)$, $e = (0, 1/2)$, $f = (1/2, 1/2)$. Moreover, you should now select some basepoint and draw the cycles around $c, d, e, f$ so that $cdef = 1 $.
Unfortunately, you can't just say that $a\mapsto CD$ and $b\mapsto DF$. While this is true on the level of homology, one has to study carefully how do the cycles with the basepoint map. I thought I'd write it now, but this takes too much time and hopefully somebody else can draw the stuff easier :)
Instead, I tried to exhibit some expressions $a, b$, which may be not exactly the cycles above, but which nevertheless satisfy $[a, b]cdef \mapsto 1$. In other words, these $a, b$ will be generators, but just different ones.
I was able to make $a = CDEC^{-1}, b = CCDC^{-1}$ work:
$$CDEC^{-1}CCDC^{-1}(CDEC^{-1})^{-1}(CCDC^{-1})^{-1}CCDDEE(CDECDE)^{-1} = $$
$$ = CDEC^{-1}CCDC^{-1}CE^{-1}D^{-1}C^{-1}CD^{-1}C^{-1}C^{-1}CCDDEE(CDECDE)^{-1} = $$
$$ = CDEC^{-1}CCDE^{-1}D^{-1}D^{-1}DDEE(CDECDE)^{-1} = $$
$$ = CDECDE (CDECDE)^{-1} = 1 $$
Finally, note that the map $ \left< C, D, E, F| CDEF = 1\right> \to \mathbb Z_2$ is given by counting all the letters modulo 2 (consistent because $F = 1 = 3 = (CDE)^{-1}$), so the image of the map discussed above should contain exactly expressions with even number of letters.