"Compression multipoint" is not common terminology. Compressing multiple points is usually done by compressing each point one after another, using point compression. That's how I'll read the question.
For the curve $y^2\equiv x^3+a\,x+b\pmod p$ with $p=23$, $a=1$, $b=1$ of that course, a method of point compression could be to express a point as $\bigl((x\bmod p),((y\bmod p)\bmod 2)\bigr)$, which is a pair of integers with the first one in $[0,p)$ and the second either $0$ or $1$. That's compressed in the sense that a single bit is used for the second integer. E.g. point $(x,y)=(7,11)$ becomes $(7,1)$
To get back at the uncompressed form, we take $x$ (here $x=7$ from the first part of the compressed form) and use the curve's equation to compute $s\gets x^3+a\,x+b\bmod p$ (here $s=7^3+7+1\bmod23=6$), then solve for $y$ the equation $y^2\equiv s\pmod p$. We can use the method there. In our example $p\bmod4=3$ thus one solution (if there is one) must be $y=s^{(p+1)/4}\bmod p$ and (if that's not $0$) the other is $p-y$. We select the appropriate $y$ thanks to the known $y\bmod 2$ (from the second part of the compressed form), then check that indeed $y^2\bmod p=s$. Here this gives $y=6^6\bmod 23=12$ or $y=23-12=11$, we select the second one since $11\bmod2=1$, and indeed $11^2\bmod23=6$. We are back at $(x,y)=(7,11)$.
For the standardized form of point compression, see this.
