To see that the diameter of this graph is always $2$, just take any two vertices $u$ and $v$ and look at the length $1$ and length $2$ paths between them.
If there is an edge $uv$, they are at distance $1$ and everything is good.
If there is no edge $uv$, then the condition $\deg(u) + \deg(v) \ge n$ applies. There are $n-2$ different possible length $2$ paths from $u$ to $v$, using the edges $ux$ and $xv$ for some third vertex $x$. Now apply the pigeonhole principle:
- There are at least $n$ edges incident to $u$ or to $v$.
- There are $n-2$ vertices $x \ne u,v$ in the graph.
- So at least $2$ of the $n$ edges must be incident to the same vertex $x$.
This gives us a path $ux, xv$ of length $2$.
You see that this proof actually works if we have $\deg(u) + \deg(v) \ge n-1$. We need $\deg(u) + \deg(v) \ge n$ for non-adjacent $u$ and $v$ for the second part of your question, which is just the statement of Ore's theorem.