Here is the fundamental polygon of the torus.
The punctured torus simply has a point $P$ removed. It's best to think of $P$ having been removed from the gray area. Your proof claims any lift of a generator of $\pi_1(U\cap V)$ is a loop (with winding number 1) around $P$. Recalling that elements of fundamental groups are invariant under homotopies, we can retract this loop to run around the boundary. Starting at the bottom right of our diagram and traveling counterclockwise, we see this look $ABA^{-1}B^{-1}$.
To be a knot in the first place, it essentially needs a well-defined regular neighborhood, sort of as proof of its tameness. If you think of a knot as being in the one-skeleton of a triangulation of $\mathbb{R}^3$, then if you want a triangulation of the complement without too many more simplices, you can remove a regular neighborhood of the knot -- after all, this is a deformation retract of $\mathbb{R}^3-K$. It is also possible to triangulate $\mathbb{R}^3-K$, but it takes infinitely many more simplices!
In any case, the van Kampen theorem applies to pairs of subcomplexes whose intersection is a path connected subcomplex. The proof involves taking a neighborhood of the intersection that deformation retracts onto the intersection, where the neighborhood is formed from neighborhoods close to faces of incident simplices, then adding this neighborhood to the pair of subcomplexes. (This is very similar to how subcomplexes of a complex form a "good pair," in Hatcher's terminology.)
There is a cell structure of the torus minus $K$ with infinitely many cells, and one can extend this to a cell structure of $\mathbb{R}^3-K$. If $X_1,X_2$ are closed sets with $X_1\cup X_2=\mathbb{R}^3$ and $X_1\cap X_2$ being the torus, then $X_1-K$ and $X_2-K$ inherit the cell structure, and the van Kampen theorem for complexes applies.
(Note: if you took the solid torus minus $K$ and the closure of the complement of this, as you suggest, their union would be all of $\mathbb{R}^3$! This is like the example of covering $S^1$ by $[0,1/2]$ and $(1/2,1]$ through the quotient map $[0,1]\to S^1$. These are two simply connected subsets of $S^1$ that intersect at a point, but van Kampen, if it were to apply, would give $\pi_1(S^1)=1$. This illustrates what open sets are meant to handle, but also $(1/2,1]$ is not a subcomplex of $S^1$.)
Best Answer
Think of this space as the closed unit square with sides properly identified, and the $(1/2,1/2)$ point removed.
Take $U$ to be an open ball around $(1/2,1/2)$. Consider a ball $B$ around $(1/2,1/2)$ that sits within $U$, and let $V$ be the complement of $B$.
So, this means that:
Because the loops in $U\cap V$ are the same loops in $U$, this means that the fundamental group is
$$\underbrace{\mathbb{Z}}_{U}\ast_{\mathbb{Z}}\underbrace{(\mathbb{Z}\ast\mathbb{Z})}_{V} \cong \mathbb{Z}\ast\mathbb{Z}.$$
You should fill in all the details, here!