Given two vector fields on a surface $v,w:S \subset \Bbb R^m \rightarrow T(S)$ my book defines the derivative of $v$ with respect to $w$ as $$\nabla_w v = (w(v_1),...,w(v_m))$$ where $(v_1,...,v_m)$ is the coordinate representation of $v$ with respect to the standard basis $\{ (\partial_{y_j})_p\}_{j=1}^m$ of the tangent space $T_p(S)$ for very $p \in S$.
These $v_i$ are functions from $S$ to $\Bbb R$ so you can apply to them the tangent vector $w_p$. So far so good.
Later though, considering a parametrization $x:U \subseteq \Bbb R^2 \rightarrow S$ it defines $x_u = \partial_u x$ and $x_v = \partial_v x$ for every $(u,v) \in U$. I consider these vector fields as the differential of $x$, i.e. the coordinates for the pushforward $(x_*)_{(u,v)}:\Bbb R^2 \rightarrow \Bbb R^m$.
Now take $x_u$. Being a vector field on $S$ we can compute $\nabla_{x_u} v$ and the book notes:
$$\nabla_{x_u} v = (\partial_u v_1,...,\partial_u v_m)$$ from which $\nabla_{x_u} x_u = x_{uu}$. I think this doesn't make sense: as I said before $v_i:S \subset \Bbb R^m \rightarrow \Bbb R$, so you can't take their derivative in $\Bbb R^2$. The second equality might be adjusted writing $\nabla_{\partial_u} x_u = x_{uu}$ since $x_u:U \rightarrow T(S) \cong \Bbb R^m$, and so is $\partial_u$. Am I right?