Fundamental group of complement of circle with a line passing through it.

Let $A=\{(x,y,z) \in X : x<\tfrac{1}{2}\}$ and $B=\{(x,y,z) \in X: x>-\tfrac{1}{2}\}$. Then $A$ and $B$ are homeomorphic and each deform retracts onto a surface $\Sigma$ surrounding "half" of the removed subspace; see below. Then $\pi_1(A)\cong \pi_1(B)\cong\pi_1(\Sigma)\cong \mathbb{Z} * \mathbb{Z} *\mathbb{Z}$. We can represent the three generators via paths as in the image below. This choice makes it clear that corresponding generators of $\Sigma_A$ and $\Sigma_B$ are equal when included into $A \cap B$. By van Kampen's Theorem, we have $\pi_1(X)\cong \mathbb{Z}* \mathbb{Z}*\mathbb{Z}$.

You ask about the Wirtinger presentations, which does extend from complements of knots in $\mathbb R^3$ to complements of graphs. This is treated in 9.1.9 of Topology and Groupoids for the case of finite graphs:

"at each vertex $v$ there is a relation $x^{\varepsilon_{1}}_{1} x^{\varepsilon_{2}}_{2}\dots x^{\varepsilon_{r}}_{r} = 1$, where the $x_{i} $ are the edges at the vertex and the sign $\varepsilon_{i}$ is $+1$ if the arrow for $x_{i}$ points towards the vertex, and $-1$ otherwise;"

For your example, here is a rough sketch:

so we get generators $a,b,c,d,e$ and relations $a=bcd, c=b^{-1}ed^{-1}$ which presents the free group on the generators $b,e,d$, which corresponds to squirrel's answer.

I think you have the right idea. Here is how I visualize what you mean:

First you shrink $\mathbb{R}^3$ down to an open ball that leaves your circle with a line through it unchanged. Since we are dealing with $\mathbb{R}^3 \setminus X $ we can think of hollowing the line and circle out a little. This gives us an open ball with a tunnel passing through from one side to the other, and it connects to the circle tunnel inside as well. Well, we can slide each half of the circle's tunnels to the outside of the ball. This will give us a ball with three tunnels passing through it, not touching each other at all on the inside. From here, we can deformation retract to a disk with three holes.

I think you know how to do it from there. Hope this helped.