There have been several questions similar to mine, but none of them fleshed out to a complete answer. I would like someone to look over this proof (I am NOT looking for alternative solutions), as it's the first time I've seriously computed with second fundamental form, Gauss equation, etc. Also, I can never read classical DG notation, so I'll stick to modern notation throughout.
I want to show the sectional curvature of the Clifford torus, that is the image of the map $F:\mathbb{R}^2\to \mathbb{R}^4$ given by $$F(\theta,\phi)=\frac{1}{\sqrt{2}}(\cos\theta,\sin\theta,\cos\phi,\sin\phi),$$ has zero sectional curvature. Let us denote the image of $F$ by $T^2$. Splitting $\mathbb{R}^4=\mathbb{R}^2_{r,\theta}\times \mathbb{R}^2_{s,\phi}$ into two copies of the plane with polar coordinates, let us take an orthonormal frame on $T^2$ as $e_1=\frac{1}{r}\partial_\theta$ and $e_2=\frac{1}{s}\partial_\phi$. Now, we compute $$sec_{(\theta,\phi)}(T^2)=R^{T^2}(e_1,e_2,e_2,e_1)=R(e_1,e_2,e_2,e_1)+g(\Pi(e_1,e_1),\Pi(e_2,e_2))-g(\Pi(e_1,e_2),\Pi(e_1,e_2))$$ where the last equality follows by Gauss equation. Since R is the curvature tensor of Euclidean space, the first term vanishes. Therefore, it remains to compute the second fundamental forms. We compute $$\Pi(e_1,e_2)=-(\nabla_{\frac{1}{r}\partial_\theta}\frac{1}{s}\partial_\phi)^\perp=-(\frac{1}{sr}\Gamma^i_{\theta\phi}\partial_i)^\perp=0,$$ where the last equality is by consideration of $g_{\mathbb{R}^4}=dr^2+r^2d\theta^2+ds^2+s^2d\phi^2$. Similarly $$\Pi(e_1,e_1)=-(\nabla_{\frac{1}{r}\partial_\theta}\frac{1}{r}\partial_\theta)^\perp=\left(\frac{\partial_r}{r}\right)^\perp=\frac{\partial_r}{r},$$ since $\partial_r$ is already perpendicular to the torus; similarly $\Pi(e_2,e_2)=\frac{\partial_s}{s}.$ But $\partial_r,\partial_s$ are orthogonal as we have a direct-sum decomposition, so all second fundamental form terms must vanish. Thus, the sectional curvature on the torus vanishes.