After watching this video, The Jacobian Matrix, I animated what I interpreted from the video
https://i.stack.imgur.com/lbL8G.jpg
While animating, I got the following doubt, when we look at $(1, 1)$, in the un-zoomed version, it's not fixed. I actually had to move the camera frame to be centered around $(1,1)$ in the zoomed in version. Now a linear transform has its origin (local $(x_0, y_0)$) fixed, I mean, $(1,1)$ should be mapped to itself if it were a linear transform, right? If the local origin is not fixed, then how do we use the Jacobian matrix to approximate the non linear transform as a linear transform locally?
Yes, it is true that the point itself is moved. But it is how neighbouring points are moved in relation to one another that the Jacobian records. So in the zoomed-in version, we are following $(1, 1)$ as it moves to $f(1, 1)$. We are keeping that point fixed in the middle of our screen, and watching how the points around $(1, 1)$ move during the same transformation. And it turns out that if we zoom in far enough, the points around $(1, 1)$ move in a fashion that looks almost linear.
More formally, if we have our function $f$, our point of interest $p$, and the Jacobian $J(p)$ at that point, then $$ f(p+v)\approx f(p) + J(p)v $$ for small vectors $v$ (where $\approx$ has a formally defined meaning). This is completely analoguous to how the regular derivative of a function works: $$ g(x + t) \approx g(x) + g'(x) t $$ for small numbers $t$ (or maybe you're more used to $g'(x)\approx \frac{g(x + t) - g(x)}{t}$, but that's the same thing). In fact, the derivative is simply the $1\times 1$ special case of the Jacobian.