$
\newcommand{\d}{\mathrm{d}\,}
\newcommand{\df}{\mathrm{d}\,f}
\newcommand{\dfx}{\mathrm{d}\,f_{\x}}
\newcommand{\x}{ \vec{\mathbf{p}}}
\newcommand{\h}{ \vec{\mathbf{x}}}
\newcommand{\p}{ \vec{\mathbf{p}}}
\newcommand{\y}{ \vec{\mathbf{y}}}
\newcommand{\xz}{ \vec{\mathbf{x}}^{0}}
$
I believe that the best way to interpret Jacobian matrix is think of it as a total derivative of a function $f:\Bbb R^n \to \Bbb R^n$.
In other words, $f' $ is the best linear approximation of $f$.
The linear approximation depends on a point at which we trying to expand $f$.
Recall the definition of differentiable function $\left( \text{ at a point } \ \x = \left[\begin{smallmatrix} x^p_1, & x^p_2, & , \dots, & x^p_n\end{smallmatrix}\right]^T\,\right)$:
$f: \Bbb R^n \to \Bbb R^n$ is called (totally) differentiable, at a point $x_0 \in\Bbb R^n $ if there exists a linear map $\dfx : \Bbb R^n \to \Bbb R^n$ such that
$$
\lim_{\x\to \y} \frac{\left\|\, f\left(\x\right) - f\left(\y\right) - \dfx \left(\x - \y\right) \,\right\|}{\left\|\, \left(\x - \y\right) \,\right\|} = 0
$$
In this case the linearization of $f$ at a point $\x$ will look like:
$$
f\left(\x + \y \right) = f\left(\x \right) + \dfx\cdot \y + R\left( \y \right)
$$
where the remainder $R\left( \y \right) = o\left( \left\|\y \right\|\right)$, i.e. $\lim_{\left\|\y \right\|\to 0} \frac{\left\|\, R\left(\y\right) \,\right\|}{\left\|\, \y \,\right\|} = 0$.
Since $\dfx: \Bbb R^n \to \Bbb R^n $ is a linear map, it can be represented as a matrix.
Using notation
$$
\h =
\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}, \qquad
f\left(\h\right) =
\begin{bmatrix}
f_1 \left(\h \right) \\
f_2 \left(\h \right) \\
\vdots \\
f_n \left(\h \right)
\end{bmatrix}
=
\begin{bmatrix}
f_1 \left(x_1, x_2, \dots, x_n \right) \\
f_2 \left(x_1, x_2, \dots, x_n \right) \\
\vdots \\
f_n \left(x_1, x_2, \dots, x_n \right)
\end{bmatrix},
$$
the total derivative of $f$ can be expressed as Jacobian matrix:
$$
\dfx =\frac{\mathrm{D}\,f\left(\h \right)}{\mathrm{D}\,\h}\bigg\lvert_{\h = \x}
=
\begin{bmatrix}
\frac{\partial f_1\left(\h \right) }{\partial x_1} \big\lvert_{\h = \x} &
\frac{\partial f_1\left(\h \right) }{\partial x_2} \big\lvert_{\h = \x} &
\cdots &
\frac{\partial f_1\left(\h \right) }{\partial x_n} \big\lvert_{\h = \x}
\\
\frac{\partial f_2\left(\h \right) }{\partial x_1} \big\lvert_{\h = \x} &
\frac{\partial f_2\left(\h \right) }{\partial x_2} \big\lvert_{\h = \x} &
\cdots &
\frac{\partial f_2\left(\h \right) }{\partial x_n} \big\lvert_{\h = \x}
\\
\vdots & \vdots & \ddots & \vdots
\\
\frac{\partial f_n\left(\h \right) }{\partial x_1} \big\lvert_{\h = \x} &
\frac{\partial f_n\left(\h \right) }{\partial x_2} \big\lvert_{\h = \x} &
\cdots &
\frac{\partial f_n\left(\h \right) }{\partial x_n} \big\lvert_{\h = \x}
\end{bmatrix}
\label{*}\tag{*}
$$
You have to keep in mind that strictly speaking the derivative $\dfx$ is defined only at a certain point $\x\in \Bbb R^n$, and that every entry in the matrix $\eqref{*}$ is a constant.
Therefore the Jacobian matrix can act on a vector $\y \in \Bbb R^n$, and the result will also be a vector in $\Bbb R^n$:
$$
\dfx \cdot \y =
\begin{bmatrix}
\frac{\partial f_1\left(\h \right) }{\partial x_1} \big\lvert_{\h = \x} &
\frac{\partial f_1\left(\h \right) }{\partial x_2} \big\lvert_{\h = \x} &
\cdots &
\frac{\partial f_1\left(\h \right) }{\partial x_n} \big\lvert_{\h = \x}
\\
\frac{\partial f_2\left(\h \right) }{\partial x_1} \big\lvert_{\h = \x} &
\frac{\partial f_2\left(\h \right) }{\partial x_2} \big\lvert_{\h = \x} &
\cdots &
\frac{\partial f_2\left(\h \right) }{\partial x_n} \big\lvert_{\h = \x}
\\
\vdots & \vdots & \ddots & \vdots
\\
\frac{\partial f_n\left(\h \right) }{\partial x_1} \big\lvert_{\h = \x} &
\frac{\partial f_n\left(\h \right) }{\partial x_2} \big\lvert_{\h = \x} &
\cdots &
\frac{\partial f_n\left(\h \right) }{\partial x_n} \big\lvert_{\h = \x}
\end{bmatrix}
\cdot
\begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix}
\implies
\dfx \cdot \y \in \Bbb R^n
$$
https://en.wikipedia.org/wiki/Total_derivative#The_total_derivative_as_a_linear_map