Thanks in advance!
You do know, I hope, that these definitions are not unique or standard.
So if we mean by a walk a sequence of alternating vertices and edges, beginning with a vertex and ending with a vertex, not necessarily the same.
A path is a walk in which all edges are distinct.
(Again these may not agree with your definitions)
So is a star-like graph with n+1 vertices, 0 to n, both “010” and “102” are walks but only “102” is a path.
Thus the number of walks of length two is $\displaystyle {\binom {n} {2}} + n$. WHY?
But the number of paths of length two is $\displaystyle {\binom {n} {2}}$. WHY?
Your interpretation is correct.
The number of walks of length two is $\displaystyle {\binom {n} {2}} + n$ because you either just choose two edges from n (all the paths) or you can walk from 0 to any vertex and back and there are n ways to do this.
and the number of paths of length two is $\displaystyle {\binom {n} {2}}$ because you just choose two edges from n and because they are all adjacent through 0 you have your path.
Please do not use PM's to ask for more help!
If n=4 then $\displaystyle Adj = \left( {\begin{array}{*{20}c} 0 & 1 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\
1 & 1 & 1 & 1 & 0 \\\end{array}} \right)$
$\displaystyle \left( {Adj} \right)^2 = \left( {\begin{array}{*{20}c}
4 & 3 & 3 & 3 & 4 \\ 3 & 4 & 3 & 3 & 4 \\ 3 & 3 & 4 & 3 & 4 \\
3 & 3 & 3 & 4 & 4 \\ 4 & 4 & 4 & 4 & 5 \\\end{array}} \right)$