I have got a differentiable mapping $f:\Bbb R^2 \to \Bbb R^2$,
Is the image of $f$ of a very small convex subset (e.g., a unit square) around any point, a bounded region?
If it is bounded, can I approximate the image as a convex hull?
My intention is to approximate a $f(x,y)$'s range locally, given a small perturbation of $\Delta x, \Delta y$ apart from $x_0, y_0$. For $R^2 \to R$ I can simply apply Taylor expansion and approximate $f$ to a linear function and then get its range as an interval. How can I deal with the 2D case?
Yes, and this is true for any continuous map in finite-dimensional spaces. The closure of bounded region is compact, the image of a compact set under a continuous map is also compact, and every compact set is bounded.
To some extent, yes. This definitely uses differentiability. On small scale, $f$ is approximately linear, which means the following: for every point $(x,y)$ and $\epsilon>0$ there is $\delta$ such that if $|\Delta x|,|\Delta y|<\delta$, then the difference between $f(x+\Delta x,y+\Delta y)$ and the linear map $f(x,y)+ A(\Delta x,\Delta y)^T$ is less than $\epsilon \delta$. Here $A$ is the matrix of derivatives of $f$.
You can still use the Taylor formula, applying it to each component of $f$. Say, $f=(u,v)$ where $u,v$ are scalar function. Approximate $u$ and $v$ separately, and you'll have an approximation for $f$.
A common practical way of approximating a smooth map is triangulation. One chooses a triangular mesh in the domain (either uniform or not, depending on how good $f$ is), and evaluates $f$ at the vertices. They draw the corresponding triangles in the target space. This does not even use derivatives, although the quality of this approximation depends on $f$ being differentiable.