Are differentials of a smooth maps tensors? Connection to Jacobian matrices?
I'm trying to learn about smooth manifolds and differential geometry in order to learn the geometric view of classical mechanics. I'm confused about what kind of "object" the differential of a smooth map is.
Set up: Let $M$ be some smooth n-dimensional manifold and let $f: M \to M $ be a smooth map from $M$ onto itself such that a point ${x}\in M$ is mapped to ${y}=f({x})\in M$. Let $(x^1,\dots ,x^n)\in \mathbb{R}^n$ and $(y^1,\dots ,y^n)\in \mathbb{R}^n$ be the coordinate representations of the points ${x}$ and ${y}=f({x})$ in some smooth chart. 916
The differential of $f$ (evaluated at ${x}$) is then a linear map between tangent spaces, $\mathbf{d} f_x: T_x M \to T_{f(x)} M \equiv T_y M$.
my question: Is it correct to say that the matrix representation of $\mathbf{d} f_x$ is given by the Jacobian $\tfrac{\partial y^i}{\partial x^j}$? If so, is $\mathbf{d} f$ a (1,1)-tensor field and $\tfrac{\partial y^i}{\partial x^j}$ are the components of $\mathbf{d} f_x \in T_y M \otimes T^*_x M$? Can we then write it using coordinate basis vector fields $\mathbf{e}_i$, and 1-forms $\boldsymbol{\epsilon}^i$ as:
$$ \mathbf{d} f _x \;=\; \frac{\partial y^i}{\partial x^j} \mathbf{e}_i|_y \otimes \boldsymbol{\epsilon}^j|_x \;=\; \frac{\partial y^i}{\partial x^j} \frac{\partial}{\partial y^i} \otimes \mathbf{d} x^j \quad \in T_y M \otimes T^*_x M \qquad (1) $$
The above is indeed a linear map $T_x M \to T_{f(x)}M$. But $\frac{\partial y^i}{\partial x^j} \frac{\partial}{\partial y^i} = \frac{\partial}{\partial x^j}=\mathbf{e}_j|_x$ so, if the above is correct (is it?), then it is the same thing as
$$ \mathbf{d} f _x \;=\; \frac{\partial}{\partial x^j} \otimes \mathbf{d} x^j \;=\; (\mathbf{e}_j\otimes \boldsymbol{\epsilon}^j)|_x \qquad\qquad (2) $$
And isn't this just the identity tensor field evaluated at the point $x$? What's going on here? Is it possible to expand the differential, either $\mathbf{d}f$ or $\mathbf{d}f_x$, as a linear combination of basis vectors (or rather, tensor products of them)? Or is my attempt to associate $\mathbf{d} f$ with a tensor misguided?
Another question: If $f$ is the map and $y^i$ are just the labels we give to its output, should the jacobian $\frac{\partial y^i}{\partial x^j}$ actually be written as $\frac{\partial f^i}{\partial x^j}$? This is perhaps a small detail with little consequence but I'm just curious. (I'm abusing notation here by using $f$ to denote both the smooth map on $M$ as well as its coordinate representation, $\varphi\circ f \circ \varphi^{-1}$, on $\mathbb{R}^n$)
Edit: question 2: Based on <PERSON> answer, I see why $\mathbf{d} f$ is not really a tensor field since $\mathbf{d} f_x$ maps between tangent spaces of two different points. 281