Assumption: $A_{i,j} = 1$ if $i\sim j$ and otherwise $A_{i,j}=0$.
Let $\Gamma_k$ denote the set of sequences in $V$ of length $k$. An element $\overline{v}\in \Gamma_k$ is of the form $\overline{v} = (v_1,\dots,v_k)$.
A path of length $k$ is an element in $\Gamma_k$ satisfying $v_{i+1} \sim v_i$ for $i=1,\dots,k-1$. A path from $i$ to $j$ of length $k$ is a path with $v_1=i$ and $v_k=j$.
A loop of length $k$ is a path of length $k$ satisfying $v_1=v_k$.
Next, by definition of matrix multiplication
$$A^k_{i,j} =\sum_{\overline{v}\in \Gamma_k:v_1=i,v_k=j}A_{v_1,v_2}A_{v_2,v_3}\dots A_{v_{k-1},v_k}$$
Also, the product on the RHS is nonzero if and only if $\overline{v}$ is a path from $i$ to $j$ of length $k$, and in this case, the product is equal to $1$.
Therefore,
$$A^k_{i,j}=\# \{\mbox{paths from }i\mbox{ to }j\mbox{ of length }k\}.$$
As a result, $\sum_{i} A^k_{i,i}$ is the number of loops of length $k$.
Yes, $Jv_i=0$ for all $i>1$. This is because, the matrix of $1$s has only one nonzero eigenvalue, $n$. Thus, since we have already covered its eigenvalue in $v_1$, and since a symmetric matrix has all independent eigenvectors, which are orthogonal to each other, therefore the vectors $v_i$ for $i>1$ are in the null space of $A(\overline{G})$, whence $Jv_i=0$.
Best Answer
Using the definition of an adjacency matrix $A$, it follows that $[A^2]_{ii}$ is equal to the number of edges connected to $i$, i.e. the degree of $i$: $d_i$. Now recall (or easily rederive) that $\sum_i d_i=2e$. Finally use the fact that the eigenvalues of $A^2$ are $\lambda_i^2$ along with the fact that their sum is equal to the trace of $A^2$.