I recently realized that I don't know any non-linear diffeomorphisms of the plane (or $\mathbb{R}^n$ in general) except for linear ones, so I want to ask rather broad questions hoping to be pointed to the appropriate literature.
1) Are there simple ways of constructing autodiffeomorphisms of $\mathbb{R}^n$ that can be expressed in closed form? UPD: Ok, there's e.g. $(x, y) \mapsto (x, y + f(x))$, where $f: \mathbb{R} \to \mathbb{R}$ is smooth, so I call this one back - kind of, if you know some exciting and unusual family, feel free to share :)
2) Is every autodiffeomorphism of $\mathbb{R}^n$ isotopic to a linear one? Obviously, every one is homotopic to any other due to $\mathbb{R}^n$ being contractible, but since $\mathrm{GL}(\mathbb{R}^n)$ is not connected, $\mathbb{R}^n$ is a k-space), and homotopies behave nicely under differentials at a point, even some linear autodiffeomorphisms are not isotopic, if I'm not mistaken.
The answer to your question (2) is yes.
The proof goes like this. Let $f$ be a diffeomorphism. We find an isotopy from $f$ to a diffeomorphism $g$ with $g(0)=0$. The isotopy is
$$(x,t) \longmapsto f(x)-tf(0) $$
when $t=0$ this is $f$, when $t=1$ you get $f(x)-f(0)$.
Next, given a diffeo $g$ with $g(0)=0$ we isotope $g$ to a linear diffeomorphism. The map is this:
$$(x,t) \longmapsto \frac{g((1-t)x)}{1-t}$$
for $t \in [0,1)$ and at $t=1$ we have
$$(x,1) \longmapsto Dg_0(x)$$
You can check this map is continuous provided $g$ is $C^1$.
Regarding your 1st question, one of the most common techniques is to integrate vector fields.