I know $\pi_1(\mathbb{R}^2\setminus\{x,y\}) = \mathbb{Z}\ast\mathbb{Z} = \langle a,b\rangle$, but its non-abelian-ness isn't obvious to me.
Specifically, I draw a box and two points to represent $x$ and $y$. I call the clockwise loop around $x$ $a$, and the clockwise loop around $y$, $b$.
If I draw the loops $ab$ and $ba$ they look `obviously homotopic' to me (they are both a single clockwise loop containing both $x$ and $y$), so should be the same element in $\pi_1$, which is clearly not the case if we take the free product of $\mathbb{Z}$ with itself.
It seems that I must be making a big mistake somewhere. What is the obstruction to $ab \simeq ba$? I would prefer something concrete, I know this fundamental group follows from standard theorems (van Kampen's theorem), but I want to see the homotopy fail, because I think they are pretty clearly homotopic - so there is something wrong with my intuition which I'd like to fix.
If they are "clearly homotopic," then you should be able to present a homotopy. If you cannot see the homotopy, then what a strange meaning you have for it being "clear"! :-)
Your winding number argument doesn't imply a homotopy, but it does end up having to do with the homology of the space, where indeed $ab$ and $ba$ are homologous paths.
I think the best way of seeing that $ab$ and $ba$ are not homotopic is through universal covering spaces or by thinking carefully about how proof of the van Kampen theorem works in this case, but instead of explaining these, I will describe an object that should exist if $ab$ and $ba$ were homotopic but whose existence seems intuitively implausible.
A homotopy $ab\simeq ba$ is equivalent to a map $f:S^1\times S^1\to \mathbb{R}^2-\{x,y\}$ where $f(t,s_0)=a(t)$ and $f(t_0,s)=b(s)$. (Take the $I\times I$ square, with $ab$ on the bottom, $ba$ on the top, and $x_0$ on the left- and right-sides. Quotient the left- and right-sides each to a point to get a function with $a$ on the bottom and top and $b$ on the sides.) That is, if you imagine a torus with $\alpha$ the major circumference and $\beta$ the minor circumference, then the map is from a torus into the twice-punctured plane mapping $\alpha$ to $a$ and $\beta$ to $b$. How exactly is the rest of this torus going to avoid $x$ and $y$? (I'm not aware of an elementary argument that gives an obstruction.)