Jacobian of the Jacobian of the function $f:\mathbb{R}^n\to\mathbb{R}^m$ Minimal Working Example
Consider the function $f:\mathbb{R}^n\to\mathbb{R}^m$. It's Jacobian function is $J_f:\mathbb{R}^n\to\mathbb{R}^{m\times n}$. What is the Jacobian of its Jacobian function?
Intuitively, it should be
$$
J_{J_f}:\mathbb{R}^{n} \to \mathbb{R}^{(m\times n) \times n }
$$
and so evaluating at a point should give $J_{J_f}(x)\in\mathbb{R}^{m\times n \times n}$. Which I guess is a rank $3$ tensor. What is the property of this tensor? Intuitively, I feel like it should be able to take two $n\times 1$ vectors and return a $m\times 1$ vector?
Context
Basically in my context I know that I need this tensor thing to have some sort of bilinearity in the sense that it takes two $n\times 1$ vectors and return a $m\times 1$ vector?
$$
J_{J_f}(a_1, a_2)[x] 
$$
Or perhaps it can take two $n\times m$ matrices and return a vector? I am so confused. The only way in which I could make it work is if it takes $n\times m \times 1$ matrices (whose transpose is has to be $(1\times n\times m)$ for things to work out) and then the dimensions work out because
$$
(1\times n\times m) \times (m\times n\times n) \times (n\times m\times 1) = 1\times n\times n\times m \times 1 = 1 \times m \times 1 = m\times 1
$$
 A: I think it is is misleading to write $J_f:\mathbb{R}^n\to\mathbb{R}^{m\times n}$. Actually the range is $M(m,n;\mathbb R)$, the vector space of all real $m \times n$-matrices. You may indentify $M(m,n;\mathbb R)$ with $\mathbb{R}^{m\times n}$, but this reveals what $J_f$ "really" does.
More formally let $\mathcal L(V,W)$ denote the set of $\mathbb R$-linear maps between real vector spaces $V, W$. It is a real vector space of dimension $\dim V \cdot \dim W$. Then the derivative of $f$ is a map
$$df : \mathbb R^n \to \mathcal L(\mathbb R^n ,\mathbb R^m) .$$
With respect to the standard bases of $\mathbb R^n ,\mathbb R^m$ the linear map $df(x) = df\mid_x$ is represented by the Jacobian matrix $J_f$ at $x$.
Now $\mathcal L(\mathbb R^n ,\mathbb R^m)$ is a finite-dimensional normed linear space. A norm can be introduced via any linear isomorphism $\phi : \mathcal L(\mathbb R^n ,\mathbb R^m) \to \mathbb{R}^{m\times n}$. But the choice of a norm is irrelevant because all norms on finite-dimensional  vector spaces are equivalent.
We can form the derivative of $df$ and get a map
$$d^2f  = ddf :  \mathbb R^n \to \mathcal L(\mathbb R^n,\mathcal L(\mathbb R^n ,\mathbb R^m)) .$$
By some elementary (multi-)linear algebra we see that $\mathcal L(\mathbb R^n,\mathcal L(\mathbb R^n ,\mathbb R^m))$ can be identified with the vector space of bilinear maps $\mathcal L(\mathbb R^n \times \mathbb R^n, \mathbb R^m)$ from
$\mathbb R^n \times \mathbb R^n$ to $\mathbb R^m$. The identification goes by associating to $L \in \mathcal L(\mathbb R^n,\mathcal L(\mathbb R^n ,\mathbb R^m))$, which is a linear map $L : \mathbb R^n \to  \mathcal L(\mathbb R^n ,\mathbb R^m)$), the bilinear map $\alpha(L) : \mathbb R^n \times \mathbb R^n \to \mathbb R^m$ given by
$$\alpha(L)(v,w) = L(v)(w) .$$
It is easily verified that $\alpha : \mathcal L(\mathbb R^n,\mathcal L(\mathbb R^n ,\mathbb R^m)) \to \mathcal L(\mathbb R^n \times \mathbb R^n, \mathbb R^m)$ is a vector space isomorphism. We can therefore regard
$$\bar d^2f = \alpha \circ d^2f :  \mathbb R^n \to L(\mathbb R^n \times \mathbb R^n, \mathbb R^m) $$
as an alternative representation of $d^2f$. You see that $\bar d^2f(x)$ is what you expected.
Iterating this, we see that we can regard the $r$-th derivative of $f$ as a map
$$\bar d^r f: \mathbb R^n \to L(\mathbb R^n \times \ldots \times \mathbb R^n, \mathbb R^m) $$
with range being the vector space of all multilinear maps from the $r$-fold product of copies of $\mathbb R^n$ to $\mathbb R^m$.
