This is a part of an exercise in Riemannian Geometry by do Carmo in page 140.
Let \begin{align*}f:\mathbb{R}^2\to \mathbb{R}^4,(\theta,\varphi)\to &~(\cos\theta,\sin\theta,\cos\varphi,\sin\varphi)\\e_1=&~(-\sin\theta,\cos\theta,0,0)\in\mathcal{X}(\mathbb{R}^4) \\e_0=&~(1,0)\in\mathcal{X}(\mathbb{R}^2)\end{align*} then $f(\mathbb{R}^2)=T^2$ is the torus and $df(e_0)=e_1,$ so $e_1$ is actually in the tangent space of $T^2$ (use the identification of $T_pT^2\subseteq T_p\mathbb{R}^4$). Now let $\overline{\nabla}$ be the Levi-Civita connection on $\mathbb{R}^4,$ I want to compute $\overline{\nabla}_{e_1}e_1,$ here is my try:
Since $e_1=-\sin\theta\frac{\partial}{\partial x^1}+\cos\theta\frac{\partial}{\partial x^2}=df(e_0),$ we have \begin{align*} \overline{\nabla}_{e_1}e_1=&~\overline{\nabla}_{df(e_{0})}\left(-\sin\theta\frac{\partial}{\partial x^1}+\cos\theta\frac{\partial}{\partial x^2}\right) \\ =&~ df(e_0)(-\sin\theta)\frac{\partial }{\partial x^1}+df(e_0)(\cos\theta)\frac{\partial}{\partial x^2}-\sin\theta\overline{\nabla}_{e_1}\left(\frac{\partial}{\partial x^1}\right)+\cos\theta\overline{\nabla}_{e_1}\left(\frac{\partial}{\partial x^2}\right) \\ =&~df(e_0)(-\sin\theta)\frac{\partial }{\partial x^1}+df(e_0)(\cos\theta)\frac{\partial}{\partial x^2} \end{align*} where the last equality follows from $$\overline{\nabla}_{e_1}\left(\frac{\partial }{\partial x^1}\right)=-\sin\theta\overline{\nabla}_{1}\frac{\partial}{\partial x_1}+\cos\theta\overline{\nabla}_{2}\frac{\partial}{\partial x_1}=0$$ so it remains to compute $df(e_0)(-\sin\theta)$ and $df(e_0)(\cos\theta),$ but if we follow the definition of differential and tangent vector, it would be $$df(e_0)(\cos\theta)=e_0(\cos f(\theta,\varphi))=\frac{\partial}{\partial\theta}\cos (f(\theta,\varphi))$$ which is incorrect since $f$ is a vector-valued function, so where do I go wrong? Is my understanding of covariant derivative incorrect? Can someone help me?