This is the easiest non trivial example I know and I consider it the most important to try to visualise and understand both Hilbert scheme of points and Hilbert-Chow morphism.
Suppose you want to understand $\rm{Hilb}^2(\mathbb{A}^2)$. First of all you may want to understand what $\rm{Sym}^2(\mathbb{A}^2)$ is. By definition $\rm{Sym}^2(\mathbb{A}^2)=\rm{Spec}\,\mathbb{C}[x,y,z,w]^{\mathfrak{G}_2}$, where $\mathfrak{G}_2=\{e,\sigma\}$ is the symmetric group with two elements and $\rm{Spec}\,\mathbb{C}[x,y,z,w]^{\mathfrak{G}_2}$ is the ring of invariant elements. Here I am assuming that you are a bit familiar with Geometric Invariant Theory, otherwise I suggest you to look at some standard references about that (Mumford, Newstead, etc..). Working and messing with this data you should finally get to the conclusion that
$$ \rm{Sym}^2(\mathbb{A}^2)=\rm{Spec}\,\mathbb{C}[a,b,c,d,e]/I$$
where $I=(e^2-abe-4cd+b^2c+a^2c)$. How visualise this? If you fix the $a$ and $b$ variables, then $\rm{Sym}^2(\mathbb{A}^2)=\rm{Spec}\,\mathbb{C}[a,b][c,d,e]/I$ where now we can recognise $I$ as the ideal of a quadric $Q\subset\mathbb{A}^3=\rm{Spec}\,\mathbb{C}[c,d,e]$ (recall that we are now supposing that $a$ and $b$ are fixed). In fact, by direct calculations, we can further say that $Q$ is actually a cone. The moral of this picture is that locally $\rm{Sym}^2(\mathbb{A}^2)$ looks like a product $\mathbb{A}^2\times Q$, where the verteces of these cones lie on the diagonal
$\Delta$ inside $(\mathbb{A}^2)^2$.
Now, in $\rm{Hilb}^2(\mathbb{A}^2)$ we have only two kind of ideals. The first kind are not so interesting, corresponding to ideals of the form $(x-a,y-b)\cap(x-a',y-b')$ where at least one between $a,a'$ and $b,b'$ is distinct, i.e. distinct points, which are sent into the point $((a,b),(a',b'))\in\rm{Sym}^2(\mathbb{A}^2)$ through the Hilbert-Chow morphism. The second kind of ideals are the ideals of length 2, which can be described in the following way. Fix a curve $C\subset\mathbb{A}^2$ and a point $p\in C$. We can consider the limit (whatever it means, if you are interested you can look at Harris-Eisenbud 'The geometry of schemes', somewhere in Chapter 2 if I am right) of another point $q\in C$ moving on $C$ and approaching $p$. This limit is the non reduced scheme whose support consists of the single point $p$, but its regular sheaf recall the direction of the crash, i.e. recall the tangent direction of the curve $C$ in the point $p$. This element will be sent in the point $(p,p)\in\Delta\subset\rm{Sym}^2(\mathbb{A}^2)$.
So we have now a good description of what $\rm{Hilb}^2(\mathbb{A}^2)$ is and how the Hilbert-Chow morphism works. To conclude we need to note that the set of tangent directions through a point is canonically identified with the projective line $\mathbb{P}^1$. The Hilbert-Chow morphism sends any ideal corresponding to one of these directions to the same point on the diagonal $\Delta$. Hence we are saying that the Hilbert-Chow morphism attach a copy of $\mathbb{P}^1$ upon each point on the diagonal (which is the singular locus of the symmetric product). Other way said, the Hilbert-Chow morphism is the blow-up of $\rm{Sym}^2(\mathbb{A}^2)$ along the diagonal.