Denote by $L^2(\mathbb{R}^n, \mathbb{R}^n)$ be the collection of of vector fields $f : \mathbb{R}^n \to \mathbb{R}^n$ such that \begin{equation} \sum_{i=1}^n \int_{\mathbb{R}^n} \lvert f_i(x) \rvert^2dx < \infty \end{equation} where each $f_i : \mathbb{R}^n \to \mathbb{R}$ is the $i^{th}$ component of $f$.
Then, $L^2(\mathbb{R}^n, \mathbb{R}^n)$ is a Hilbert space. Moreover, the $2-$fold tensor product of $L^2(\mathbb{R}^n, \mathbb{R}^n)$ is another Hilbert space generated by the following set \begin{equation} \{ \{f_i(x) g_j(y) \} \mid f, g \in L^2(\mathbb{R}^n, \mathbb{R}^n)\} \end{equation} with respect to the inner product \begin{equation} \Bigl \langle \{f_i(x) g_j(y) \} , \{h_i(x) w_j(y) \} \Bigr \rangle:= \sum_{i,j=1}^n \Bigl[\int_{\mathbb{R}^n} f_i(x) h_i(x) dx\Bigr]\Bigl[\int_{\mathbb{R}^n} g_j(y) w_j(y) dy\Bigr]. \end{equation}
However, in the context of differential geometry, I see that given a smooth manifold $M$ and two smooth vector fields $X,Y$ on it, the tensor produt $X \otimes Y$ is defined as \begin{equation} X \otimes Y : M \to TM \otimes TM \text{ where } (X \otimes Y)_p:=X_p \otimes X_p \in T_pM \otimes T_pM \text{ for each } p \in M. \end{equation}
This causes extreme confusion for me...For the Hilbert space, $f$ and $g$ have "different arguments" when forming a tensor product: $f\otimes g= \{ f_i(x) g_j(y) \}$.
But, for differential geometry, $X \otimes Y$ is still evaluated at "a common point": $(X\otimes Y)_p$
Could anyone please clarify??