Here is an elementary approach. $\#(M\cap M_i)$ equals the number of solutions $y-x=i$ with $x,y\in M$. Every quadratic residue is the square of two nonzero residues, so $\#(M\cap M_i)$ equals $1/4$ times the number of solutions of $a^2-b^2=i$ with $a,b\in\mathbb{F}_p^\times$. For $p>2$ we can make the bijective change of variable $u:=a+b$ and $v:=a-b$, which shows that we are counting the number of solutions $uv=i$ with $u,v\in\mathbb{F}_p$ such that $u\neq\pm v$. Without the restriction $u\neq\pm v$, there are exactly $p-1$ solutions. If $i$ is a quadratic residue, then there are precisely $4$ solutions with $u=\pm v$, otherwise there are no such solutions. 

To summarize, $\#(M\cap M_i)$ equals $\frac{p-5}{4}$ or $\frac{p-1}{4}$ depending on whether $i$ is a quadratic residue or not.