Suppose $f:\mathbb R^2\to \mathbb R$ is defined by $f(x,y)=x^2+xy$.
The gradient is $(2x+y, x)$ and so the Jacobian is $\begin{bmatrix} 2x+y& x\end{bmatrix}$.
At the point $(1,2)$, the derivative is the linear transformation $\begin{bmatrix} 4& 1\end{bmatrix}$, which is the matrix of the transformation $T(x,y)=4x+y$.
For the sake of an example, the directional derivative in the direction of $(1,1)$ is $\begin{bmatrix} 4& 1\end{bmatrix}\begin{bmatrix} 1/\sqrt{2}\\ 1/\sqrt{2}\end{bmatrix}=5/\sqrt{2}$. Which I understand is the slope of the tangent plane in the direction of $(1,1)$.
However, I am not quite understanding what the linear transformation $T(x,y)=4x+y$ represented by the matrix $\begin{bmatrix} 4& 1\end{bmatrix}$ is supposed to represent.
The tangent plane lives tangent to the graph of $f$ at the point $f(1,2)=3$; i.e., at the point $(1,2,3) \in \mathbb R^3$.
So, what exactly does the linear transformation $\begin{bmatrix} 4& 1\end{bmatrix}$ represent?
What is being transformed?
In two dimensions, you can move in a circles's worth of directions. These directions are taken in a tangent space around the point at which the derivative is calculated. In your example that is the point $(1,2)$.
The derivative $[4,1]$ means that if I move by an amount $(dx_1, dx_2)$ from $(1,2)$ then I will move $dy = 4dx_1+dx_2$ from $3$ which is $f(1,2)$.
In general, if $f:\mathbb R^m\to\mathbb R^n$ is differentiable at $(x_1,...,x_m)$ then $Df(x_1,...,x_m)$ is a linear map that maps $(dx_1,...,dx_m)$ to $(dy_1,...,dy_n)$ where $(dx_1,...,dx_m)$ is in the tangent space of $(x_1,...,x_m)$ and $(dy_1,...,dy_n)$ is in the tangent space of $f(x_1,...,x_m)$.