You are incorrect in your calculation of $\pi_1(U_g)$; it's free on $2g+1$ generators. What you need now is that the map $U_g \to U_{g+1}$ induces the map $F_{2g+1} \to F_{2g+3}$ sending the $i$th generator of the first to the $i$th generator of the second. Now we have...
If a space $X$ is the union of a directed set of subspaces $X_\alpha$ ordered by inclusion with the property that each compact set in $X$ is contained in some $X_\alpha$, and each $X_\alpha$ contains a given point $x_0$, then the natural map $\varinjlim \pi_1(X_\alpha,x_0) \to \pi_1(X,x_0)$ is an isomorphism.
Surjectivity follows from the compactness hypothesis; the image of a representing map $f: S^1 \to X$ of an element of $\pi_1(X,x_0)$ lies in some $X_\alpha$, so is in the image of some $f' \in \varinjlim\pi_1(X_\alpha, x_0)$. Suppose $f \in \varinjlim \pi_1(X_\alpha, x_0)$ maps to zero; then we have a null-homotopy $f_t: S^1 \times I \to X$. Since the image of this is compact, this defines a null-homotopy $f_t: S^1 \times I \to X_\alpha$ for some $\alpha$, so $f$ is zero in $\varinjlim \pi_1(X_\alpha, x_0)$ as well.
(This fact, and its proof, is an adaptation of Proposition 3.33 in Hatcher's algebraic topology book.)
Taking the limit of $\pi_1(U_g,x_0)$ gives us that $\pi_1(U) \cong F_\infty$, the free group on countably many generators. Similarly for $\pi_1(V,x_0)$.
Now $U \cap V \cong S^1$, and the inclusion $S^1 \to U, S^1 \to V$ induces the map $F_1 \to F_\infty$ sending the generator of the first to a chosen generator of the second. Thus by the van Kampen theorem, $$\pi_1(U \cup V, x_0) \cong \langle u_1, v_1, u_2, v_2, \dots \, \mid \, u_1 = v_1\rangle,$$
giving $\pi_1(X,x_0) = \pi_1(U \cup V, x_0) \cong F_\infty$ as desired.
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
Contrary to what you state, the count of vertices of the octagon quotient (namely the quotient space obtained by identifying opposite sides of the octagon) depends very much on the orientations of the sides.
You can see this by writing out possible "gluing words".
For example, using the gluing word $abcda^{-1}b^{-1}c^{-1}d^{-1}$ it does follow that the octagon quotient has only 1 vertex and therefore has Euler characteristic equal to $1-4+1=-2$. With more work you can see that this surface is indeed homeomorphic to $T_2$.
But consider next the gluing word $abcdabcd$. Using this you will see that the octagon quotient has $4$ vertices, not $1$ vertex. The Euler characteristic is therefore equal to $4-4+1=1$. So this octagon quotient certainly is not homeomorphic to $T_2$ nor to $P_4$. I'll leave it to you to work which which of the $T_n$'s or $P_m$'s this surface is homeomorphic to, and then to continue your exercise by calculating all the other possibilities.