I know by Hopf classification theorem that $[T^2;S^2]$(torus to sphere) are classified by the integral cohomology group $H^2(T^2;\mathbb{Z})\approx\mathbb{Z}$. Also I understand that in general, given any continuous map $\phi:T^2\to S^2$, we can first obtain a cellular approximation $\tilde{\phi}$ of $\phi$, then the problem reduces to simple cellular decompositions and calculations of degrees of maps on cells.
The problem is that the above procedure seems very hard to be programmed into a computer. I'm wondering, if I only care about smooth maps, is there a calculus-theoretic formula of doing it? Let's, say use the following notations, $$\phi:(z_1, z_2)\mapsto(u(z_1,z_2),v(z_1,z_2),w(z_1,z_2)),$$ where $(z_1, z_2)=(\exp(ix_1), \exp(ix_2)), x_1,x_2\in[0,2\pi)$ and $u^2+v^2+w^2=1$; or any other notation you find more convenient.
You're looking for calculus, so it's very simple. You want to take a $2$-form that generates $H^2(S^2)$, pull it back by your mapping, and integrate over the torus. So, if you had the mapping in spherical coordinates, most conveniently, say $$f(e^{2\pi i s},e^{2\pi i t}) = (\phi(s,t),\theta(s,t)),$$ then you would pull back $\frac1{4\pi}\sin\phi d\phi d\theta$, which is the area $2$-form, integrate $ds\,dt$ over $[0,1]\times [0,1]$, and that would be your degree.
(If your mapping is, as you indicated, given in cartesian coordinates, then just pull back $\frac1{4\pi}(x dy\wedge dz + y dz\wedge dx + z dx\wedge dy)$ and integrate.)
If you need further clarification, let me know.