For any cardinal $\kappa$ and any totally ordered set $(S,\le)$ such that $|S| > 2^\kappa$, does $S$ necessarily have at least one subset $T$ such that either $\le$ or its opposite order $\ge$ well-orders $T$ with an order type equal to the initial ordinal of $\kappa^+$?
Equivalently, for any infinite cardinal $\kappa$ and any totally ordered set $(S,\le)$ such that every subset has both a cofinality and a coinitiality no greater than $\kappa$, does it hold that $|S|\le 2^\kappa$?
If not, does there exist a $\kappa\ge\aleph_0$ such that it does hold in the special case of that specific $\kappa$?
Yes, any totally ordered set $S$ of cardinality $|S|\gt2^\kappa$ contains either a subset of order type $\kappa^+$ or else a subset of order type $(\kappa^+)^*$ (the reverse of $\kappa^+$).
Proof. We identify cardinal numbers with their initial ordinals. Consider an injection $f:(2^\kappa)^+\to S$. By the Erdős–Rado theorem there is a subset of $(2^\kappa)^+$ of cardinality $\kappa^+$ on which $f$ is either strictly increasing or else strictly decreasing.