We have the two-dimensional torus $T^2$ parametrized by $F(a,b)=(e^{ia}, e^{ib})\subset \mathbb{R}^4$ (or $\mathbb{C}^2$). Therefore, there is a Riemannian metric on $T^2$ via the pull-back from the Riemannian metric on $\mathbb{R}^4$. How would one go about computing $[\frac{\partial}{\partial \alpha}, \frac{\partial}{\partial\beta}]$ and $\Delta_{\frac{\partial}{\partial \alpha}}\frac{\partial}{\partial \beta}$ but I am not sure where to start with this.
For the commutator, I know $[X,Y]=XY-YX$, so wouldn't you just have $[\frac{\partial}{\partial \alpha}, \frac{\partial}{\partial\beta}]f=0$ for any $f\in C^\infty(T^2)$ since the mixed partials commute? I am not sure if this is correct or how to proceed.
Coordinate vector fields always commute since mixed partial derivatives commute. This is trivial and a general fact true for any manifold. For the rest of the geometry of $T^2$, it might be easier to use some more standard notation: consider the parametrization $X\colon ]0,2\pi[^2 \to T^2$ given by $$X(u,v) = (\cos u, \sin u, \cos v, \sin v).$$Then $$\frac{\partial}{\partial u}\bigg|_{X(u,v)} =X_u(u,v) = -\sin u\frac{\partial}{\partial x}\bigg|_{X(u,v)} + \cos u\frac{\partial}{\partial y}\bigg|_{X(u,v)}$$and $$\frac{\partial}{\partial v}\bigg|_{X(u,v)} =X_v(u,v) = -\sin v\frac{\partial}{\partial z}\bigg|_{X(u,v)} + \cos v\frac{\partial}{\partial w}\bigg|_{X(u,v)}.$$The metric in $T^2$ induced from $\Bbb R^4$ expressed in the coordinates $(u,v)$ is just ${\rm d}s^2 = {\rm d}u^2 + {\rm d}v^2$. In particular, it follows that all the Christoffel Symbols vanish, and that the metric is flat. So, if $\nabla$ denotes the Levi-Civita connection of $T^2$, we have that $$\nabla_{\partial/\partial u}(\partial/\partial u) = \nabla_{\partial/\partial u}(\partial/\partial v) = \nabla_{\partial/\partial v}(\partial/\partial v) = 0,$$and the (vector-valued) second fundamental form $\alpha$ of $T^2$ is given by the second derivatives of $X$: $$\begin{align}\alpha\left(\frac{\partial}{\partial u}, \frac{\partial}{\partial u}\right) &= X_{uu} = -\cos u\frac{\partial}{\partial x} - \sin u\frac{\partial}{\partial y} \\ \alpha\left(\frac{\partial}{\partial u}, \frac{\partial}{\partial v}\right) &= X_{uv} = 0 \\ \alpha\left(\frac{\partial}{\partial v}, \frac{\partial}{\partial v}\right) &= X_{vv} = -\cos v\frac{\partial}{\partial z} - \sin v\frac{\partial}{\partial w}\end{align}$$The mean curvature is $$2H = \alpha(\partial_u,\partial_u)+\alpha(\partial_v,\partial_v) = -X,$$i.e., for all $p \in T^2$ we have $H(p) = -p/2$.