I believe the answer to the first question is yes. Since ordered algebraic extensions of $\mathbb Q$ are archimedean, all orderings on a number field are induced by its real places, and in particular, there are only finitely many. Thus, there is a finite set $\{a_1,\dots,a_n\}\subseteq F$ such that every order is uniquely determined by the signs of $a_1,\dots,a_n$. For a given ordering $<$, let $E\subseteq\mathrm{rcl}(F,<)$ be an extension of $F$ containing $\sqrt{a_i}$ for $a_i>0$, and $\sqrt{-a_i}$ for $a_i<0$. Then every order on $E$ extends $<$, which implies that every $<$-positive element of $F$ is a sum of squares in $E$.

I believe the answer to the first question is yes. Since ordered algebraic extensions of $\mathbb Q$ are archimedean, all orderings on a number field are induced by its real places, and in particular, there are only finitely many. Thus, there is a finite set $\{a_1,\dots,a_n\}\subseteq F$ such that every order is uniquely determined by the signs of $a_1,\dots,a_n$. For a given ordering $<$, let $E\subseteq\mathrm{rcl}(F,<)$ be an extension of $F$ containing $\sqrt{a_i}$ for $a_i>0$, and $\sqrt{-a_i}$ for $a_i<0$. Then every order on $E$ extends $<$, which implies that every $<$-positive element of $F$ is a sum of squares in $E$.

Also, if I am to interpret the second question as “do there exist ordered number fields other than $\mathbb Q$ that force their own order”, the correspondence of orderings with real embeddings shows that this happens if and only if the field has a unique real embedding. Examples abound, such as $\mathbb Q(\sqrt[3]2)$.