Here's a bit of an elaboration on my earlier comment.
You can see $SO_3$ as the quotient of a unit $3$-ball by the antipodal map on the boundary. This is realized explicitly by the exponential map for $SO_3$. The tangent space to the identity of $SO_3$ is all $3\times 3$ matrices $H$ such that $H^T=-H$. Up to conjugation by an element of $SO_3$, such a matrix looks like
$$\left(\begin{matrix} 0 & \theta & 0 \\ -\theta & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\right)$$
which under exponentiation is converted into rotation by $\theta$ about the $z$-axis. Every element of $SO_3$ is rotation by some angle about a fixed axis since the characteristic polynomial has $1$ as a root -- so this is a rather explicit description of how the exponential map is onto. If you choose the length of a matrix $[a_{i j}]$ to be $\sqrt{ \sum_{i j} a_{i j}^2}$, then technically $SO_3$ is the quotient of the ball of radius $\pi/\sqrt{2}$, with antipodal points on the boundary identified. But that's small potatoes.
The 3-ball with antipodal points identified, as you've seen has fundamental group $\Bbb Z_2$. In particular you can think of the 3-ball with antipodal points identified as being built up from the 1-ball with antipodal points identified, union a 2-cell (this gives a CW-decomposition of the 2-ball with antipodal points identified) then union a 3-cell. So the fundamental group all "happens" in the 2-ball with antipodal points identified. The 1-skeleton generates the fundamental group and in our model (under exponentiation) corresponds to one full rotation by $2\pi$ about any chosen axis. The relation corresponds to the 2-cell attachment and you can literally interpret it as the plate trick in a "sufficiently nice" model. Let me explain.
The trouble with the plate trick is it uses a human arm. Human arms you can think of as carrying a core underlying piece of information, a map $f : [0,1] \to SO_3$. $f(t)$ represents the orientation in space of the part of the arm distance $t$ from the shoulder, so $0$ represents the shoulder distance, and $1$ the palm of the hand. There are technical issues -- if you choose the framing corresponding to the positions of the appropriate bone in the arm, you don't get a continuous map, just a piecewise-constant map -- but if you consider the orientations of the ligaments between the bones it becomes a continuous map. So the trouble with the plate trick is the map $f$ isn't the only constraint on the arm. The arm is embedded in space. Bones are rigid, there's only so much ligaments are capable of doing. So you could be rightly concerned that the plate trick is a demonstration of a peculiarity of anatomy, rather than something about $SO_3$.
Alright, but formally what's going on with the plate trick? As mentioned, a configuration of the arm we're representing by $f$, which is an element of the path space $PSO_3$ of continuous paths in $SO_3$ with $f(0) = I \in SO_3$ fixed.
There is the path-loop fibration:
$$\Omega SO_3 \to PSO_3 \to SO_3$$
and $PSO_3$ is contractible. The plate trick "is" the induced map $\pi_1 SO_3 \to \pi_0 \Omega SO_3$ from the homotopy long exact sequence of the above fibration, which is guaranteed to be an isomorphism by the path-loop fibration. Specifically, elements of $\pi_1 SO_3$ you are thinking of as motions of your hand. The path-loop fibration is telling you that you can realize the motion as a motion of your arm (or at least, a path in $PSO_3$ but the fact that people can do the plate trick says you can actually realize it as a motion of your arm).
There is of course an element that's a bit misleading in all this. The arm isn't complicated enough to allow for you to lift the product of two generators of $\pi_1 SO_3$ and witness the null homotopy -- since the arm allows a fixed complexity, the null-homotopy happens as you lift the concatenation.
But you can see from the rather explicit exponential map above how the null homotopy actually works, as it's sitting in the image of the 2-ball with antipodal points identified.
How is that?
Recall:
Fact: Let $G$ be a topological group. For any neighbourhood $U$ of $1$, there exists a (possibly smaller) open neighbourhood $V\ni 1$ such that $V=V^{-1}$ and $V^2\subseteq U$.
Proof: Let $V=U_1\cap U_2\cap U_1^{-1}\cap U_2^{-1}$, where $U_1\times U_2\subseteq m^{-1}(U)$ is a neighbourhood of $(1,1)$, $m\colon G\times G\to G$ is the multiplication in $G$.
Now for your question, note that WLOG we may assume $x=1$.
Take the open neighbourhood $U=M-Gy$ (note $Gy$ is closed since $M$ is Hausdorff) of $1$ and obtain $V\ni 1$ with $V^2\subseteq U$ and $V=V^{-1}$.
We claim the neighbourhoods $V$ of $x$ and $yV$ of $y$ works.
Indeed, if $z\in gV\cap yV$, then $g^{-1}z\in V$ and $y^{-1}z\in V$, so $g^{-1}y=(g^{-1}z)(y^{-1}z)^{-1}\in VV^{-1}=V^2$, contradicting the construction $V^2\subseteq M-Gy$.
Best Answer
A 360 degree rotation of the arm results in a very sore arm unless you unwind it or wind it a second time. Think about the rotation of the hand as being a path in the space of all rotations of 3-space. Or if you really want to rotate something, tape a belt to a soccer ball and tape the other end of the belt to the wall. Just rotate the ball a full rotation, the belt gets a twist. Now rotate again; the belt has two full twists. The twists in the belt can be undone by looping the ball back through the belt as in the animation.
The diagram in the lower right shows the paths in the space $\operatorname{SO}(3)$ which is the $3$-ball with its antipodal points on its boundary identified. That is why the doubled path appears to be broken; it is passing through antipodal points.
You might play with a Möbius band and the pencil loop which maps twice around to gain more intuition. The point of thinking of the Möbius band is that it is embedded in the space of rotations.
It is really worth thinking this through while going to sleep or while riding the bus!