We will consider $$ \pi_1 X,\ X:=\mathbb{R}^4-P_1-P_2$$ in terms of two-convexity notion (which will be addressed in the following question), where $$P_1=\{(x,y,0,0)\in\mathbb{R}^4\},\ P_2=\{(0,0,z,t)\in\mathbb{R}^4\}$$ are coordinate planes.
Question : If $P$ is any plane in $\mathbb{R}^4$ and $c$ is a closed curve in $P\cap X$, and if $[c]$ is $0$ in $\pi_1(X)$, then $[c]$ may not be $0$ in $\pi_1(P\cap X)$. Hence $X$ is not two convex
Proof : Note $\pi_1X=\pi_1(S^3-S_1-S_2)$ where $S_i$ are disjoint great circles in $S^3$. Assume that $P\cap S^3$ is a circle $c$. Nontrivial case is the case where $c$ and $S_1$ are linked. But I can not proceed any more.
Thank you in advance.
[Add] There is interpretation of two-convexity : If $U$ is $r$-tubular neighborhood of $P_1\cup P_2$, i.e. set of points whose distance from $P_1\cup P_2$ is less than $r$, then consider $\partial U$. If we perturb $\partial U$, then we have a smooth 3-dimensional submanifold $S$.
EXE : Two-convexity iff at most one principle curvature is negative.
Rough Proof : $\Leftarrow$ : If $P$ is a plane, then $P\cap S$ is a closed curve $c=\partial D$ where $D\subset S$ is two dimensional disk. If $c$ is not homologous to a point in $P\cap S$, then there are two negative principle curvatures on $S$.
You actually have the whole solution right there, but your trivial case where $c$ and $S_1$ are unlinked is really the only case with merit.
As you stated, for $P_1 = \{(x,y,0,0) \in \mathbb R\}$, $P_2 = \{(0,0,z,t)\in \mathbb R\}$; $X = \mathbb R - P_1 - P_2$ deformation retracts onto $S^3 - S_1 - S_2 = Y$ where $S_1,S_2$ are disjoint (linked) great circles in $S^3$. Likewise the plane $P$ corresponds to a circle $c$ in $S^3$. If $c$ were unlinked from $S_1,S_2$ then any closed curve on $c$ (id est an integer multiple of $c$), is trivial in $Y$ but certainly not trivial in $Y \cap P = P$. For the case where $c$ is linked, any closed curved on $c$ will not be trivial in $Y$ and the statement is vacuously true.