What exactly does the belt/plate trick demonstrate?
Stretch your arm out.
Rotate your hand as in the plate trick. Start at time $t=0$, complete one rotation at time $t=1$, and complete the second rotation at time $t=2$.
Think of your arm as a copy of the unit interval, stretching from $s=0$ at your shoulder to $s=1$ at the tip of your hand.
Imagine a copy of the standard basis for ${\mathbb R}^3$ attached to each point of your arm at the beginning of the motion. A rotation of that basis is an element of $SO(3)$. At time $t$, the basis at point $s$ has undergone some rotation. So we have a map $f:(s,t)\mapsto SO(3)$.
This graph will help you visualize that map:
Note that at the end of one rotation, the tip of your hand ($s=1$) has returned to its original orientation. Therefore $f(1,0)=f(1,1)=f(1,2)$. This means that the map $t\mapsto f(1,t)$ is a closed loop in $SO(3)$, the image of the upper edge of the left-square --- and also the image of the upper edge of the right square, because the second rotation is identical to the first. Call this loop $Q$. It's the loop traced out over time by the basis at the tip of your hand.
Note that at time $t=1$, the tip of your hand is restored to its original orientation, as is your shoulder (which in fact has maintained its orientation throughout). Therefore $f(0,1)=f(1,1)$, so that the map $s\mapsto f(s,1)$ is another closed loop. It's the loop traced out along your arm at the end of one rotation, starting from the trivial rotation at your shoulder and returning to the trivial rotation at your fingertip. Call this loop $P$.
Theorem: The paths $P$ and $Q$ are homotopic (that is, one can be deformed continuously into the other). Proof: Consider Square A. Alternate proof: Consider Square B.
Theorem: The path $P$ cannot be homotopic to a constant loop. Proof: Any homotopy would give a way of untwisting your arm without changing the orientation of your fingertip along the way. This is evidently impossible.
Theorem: The path $Q+Q$ (that is, two full rotations of your arm) is homotopic to a constant loop. Proof: Consider squares $A+B$. The trivial paths along the vertical edges tell you that this rectangle is a homotopy from Q+Q to the constant loop, keeping the endpoints fixed along the way.
Main theorem: $\pi_1(SO(3))$ contains an element of order $2$. Proof: By (8) and (9), $Q$ is not homotopic to a constant loop, i.e. $[Q]\neq 0$. But by (10), $2[Q]=0$.
In particular, $SO(3)$ is not simply connected and therefore not equal to its own universal cover (which is of course $SU(2)$).
Note: This does NOT seem to me to prove that $\pi_1(SO(3))$ has exactly two elements; only that it has at least two, and that one of those has order two. I have sometimes heard it claimed that the plate trick actually nails down the full structure of $\pi_1(SO(3))$ but I'll be very surprised (and delighted!) if someone can make that argument.