Please check carefully.

Denote $d_i=\deg(v_i)$, $\delta_i=\deg(w_i)$. We assume (without loss of generality) that $d_1\geqslant d_2\geqslant\ldots \geqslant d_n$ and $\delta_1\geqslant \delta_2\geqslant\ldots \geqslant \delta_n$. We start with the observation similar to that of Erdős, Chen, Rousseau and Schelp: there exists $s\in \{1,2,\ldots,n-k+1\}$ such that $d_s\leqslant \delta_{s+k-1}+k-1$.

Indeed, assume the contrary: $d_i\geqslant \delta_{i+k-1}+k$ for all $i=1,2,\ldots,n-k+1$. Then the total number of edges incident to $v_1,\ldots,v_{n-k+1}$ is at least $\delta_k+\ldots+\delta_n+k(n-k+1)$. Thus at least $k(n-k+1)$ these edges go to the vertices $\delta_1,\ldots,\delta_{k-1}$ (call such edges interesting), but each of them may be incident to at most $n-k+1$ interesting edges, so totally we have at most $(k-1)(n-k+1)$ interesting edges. A contradiction.

Analogously, there exists $t\in \{1,2,\ldots,n-k+1\}$ such that $\delta_t\leqslant d_{t+k-1}+k-1$.

Call $(i,j)$ a $D$-pair, corr. $\Delta$-pair, if $d_i\geqslant \delta_j+k$, corr. $\delta_j\geqslant d_i+k$. Since $k>n/2$, we have $d_i>n/2$ for a $D$-pair $(i,j)$ and $d_i<n/2$ for a $\Delta$-pair $(i,j)$. Thus there exists $i_0\in \{1,2,\ldots,n+1\}$ such that $i\leqslant i_0-1$ for any $D$-pair $(i,j)$, and $i\geqslant i_0$ for any $\Delta$-pair $(i,j)$. Analogously, choose $j_0\in \{1,2,\ldots,n+1\}$ such that $j\leqslant j_0-1$ for any $\Delta$-pair $(i,j)$, and $j\geqslant j_0$ for any $D$-pair $(i,j)$. Another restriction for a $D$-pair $(i,j)$ is that either $i<s$ or $j>s+k-1$ (or both). Analogously, for $\Delta$-pairs we have $j<t$ or $i>t+k-1$. And having all these restrictions (now forget about the initial graph), I claim that the total number of $D$-pairs and $\Delta$-pairs is at most $2k(n-k)$.

The proof is bit boring. Fix $i_0$ and $j_0$ and look for the value of $s$ for which the number of $D$-pairs satisfying our restrictions is maximal possible. It is not hard to see that it is maximized when $s=1$ or $s=n-k+1$ (if $i_0\leqslant n-k+1$ then putting $s=n-k+1$ we get no extra restrictions from "$i<s$ or $j>s+k-1$"; analogously, if $j_0\geqslant k+1$ we get no extra restriction for $s=1$. Finally, if $i_0\geqslant n-k+2$ and $j_0\leqslant k$, the extra restriction remove $(i_0-s)(s+k-j_0)$ pairs, which is a concave function maximized on the interval $s\in [1,n-k+1]$ in one of the endpoints). Hence we may assume $s=1$ or $s=n-k+1$, analogously $t=1$ or $t=n-k+1$. There are two essentially different cases now:

1. $s=1$, $t=n-k+1$. We get $j\geqslant k+1$ for $D$-pairs and $j\leqslant n-k$ for $\Delta$-pairs $(i,j)$. Thus any $i$ participates in at most $n-k$ $D$-pairs and in at most $n-k$ $\Delta$-pairs, totally $i$ participates in at most $n-k$ pairs (since it can not participate in both types of pairs). Therefore the total number of pairs does not exceed $n(n-k)\leqslant 2k(n-k)$.

2. $s=t=1$. We get $j\geqslant k+1$ for $D$-pairs and $i\geqslant k+1$ for $\Delta$-pairs $(i,j)$. Denote $a=\max(k+1,j_0)$, $b=\max(k+1,i_0)$. Then the total number $D$-pairs is at most $(n-a+1)(b-1)$, the total number $\Delta$-pairs is at most $(n-b+1)(a-1)$. When $a,b\in [k+1,n+1]$, the sum $S:=(n-a+1)(b-1)+(n-b+1)(a-1)$ which is bilinear in $a$ and $b$ is maximized at one of four endpoints. When $a=b=k+1$ we get $S=2k(n-k)$, if, say, $a=n+1$, we get $S=n(n-b+1)\leqslant n(n-k)\leqslant 2k(n-k)$.