I want to show that if I remove $n$ parallel lines from $\mathbb{R}^3$ then I get $\mathbb{R}^2\setminus \{p_1,\dots,p_n\}.$
There is also some underlying structure I wish to also understand. That is, $\pi_1(x_0,\mathbb{R}^2\setminus\{p\})$ is $\mathbb{Z}$. Does this generalize? More precisely, is $\pi_1(x_0,\mathbb{R}^2\setminus\{p_1,\dots,p_n\}) =\mathbb{Z}^n$?
Call the lines $L_1, \dots, L_n$.
Let $w$ be a direction vector for the lines, and let $P$ be the plane through the origin with normal vector $w$. Note, all the lines are orthogonal to $P$.
Let $\{u, v\}$ be a basis for $P$, then $\{u, v, w\}$ is a basis for $\mathbb{R}^3$. Define the linear transformation $T : \mathbb{R}^3 \to \mathbb{R}^3$ by $T(u) = e_1$, $T(v) = e_2$, and $T(w) = e_3$. Note that $T$ sends $P$ to the $xy$-plane, and the lines orthogonal to $P$ to lines orthogonal to the $xy$-plane (i.e. lines with direction vector $e_3$); let $\ell_i = T(L_i)$. As $T$ is invertible, it is a homeomorphism so $\mathbb{R}^3\setminus\{L_1, \dots, L_n\}$ is homotopy equivalent to $\mathbb{R}^3\setminus\{\ell_1, \dots, \ell_n\}$.
If $\pi$ denotes the projection $\mathbb{R}^3 \to \mathbb{R}^2$ given by $\pi(x, y, z) = (x, y)$, let $p_i$ be the projection under $\pi$ of the point at which $\ell_i$ intersects the $xy$-plane. Then $\pi$ restricts to a map $\pi : \mathbb{R}^3\setminus\{\ell_1, \dots, \ell_n\} \to \mathbb{R}^2\setminus\{p_1, \dots, p_n\}$. We also have the inclusion $i : \mathbb{R}^2 \to \mathbb{R}^3$ given by $i(x, y) = (x, y, 0)$. Note that $i$ restricts to a map $i : \mathbb{R}^2\setminus\{p_1, \dots, p_n\} \to \mathbb{R}^3\setminus\{\ell_1, \dots, \ell_n\}$.
While $\pi\circ i : \mathbb{R}^2\setminus\{p_1, \dots, p_n\} \to \mathbb{R}^2\setminus\{p_1, \dots, p_n\}$ is the identity map, $i\circ\pi : \mathbb{R}^3\setminus\{\ell_1, \dots, \ell_n\} \to \mathbb{R}^3\setminus\{\ell_1, \dots, \ell_n\}$ is given by $(i\circ\pi)(x, y, z) = (x, y, 0)$. Note however that $i\circ\pi$ is homotopic to the identity map via the homotopy
\begin{align*} H : (\mathbb{R}^3\setminus\{\ell_1, \dots, \ell_n\})\times[0, 1] &\to \mathbb{R}^3\setminus\{\ell_1, \dots, \ell_n\}\\ H((x, y, z), t) &= (x, y, tz). \end{align*}
Therefore, $\pi$ and $i$ are homotopy inverses, so $\mathbb{R}^3\setminus\{\ell_1, \dots, \ell_n\}$ and $\mathbb{R}^2\setminus\{p_1, \dots, p_n\}$ are homotopy equivalent. As homotopy equivalence is transitive, $\mathbb{R}^3\setminus\{L_1, \dots, L_n\}$ and $\mathbb{R}^2\setminus\{p_1, \dots, p_n\}$ are homotopy equivalent.
As for your question about fundamental groups, note that $\mathbb{R}^2\setminus\{p_1, \dots, p_n\}$ is homotopy equivalent to a bouquet of $n$ circles, the fundamental group of which is $F_n$, the free group on $n$ generators. Note, this is not the same group as $\mathbb{Z}^n$ unless $n = 1$. For example, if $n > 1$, $F_n$ is not abelian while $\mathbb{Z}^n$ is.