If $G$ is a group with a (left) linear order, does every (left) partial order on $G$ extend to a (left) linear order? 


The answer is affirmative on abelian groups, where being torsion-free is necessary and sufficient both for having a linear order and for partial orders to extend to linear orders ([Fuchs, 1950](http://www.jstor.org/stable/2372145), since any partial order on a torsion-free abelian group extends to a normal order in his terminology).



Extensions of partial orders to linear orders on (nonabelian) groups
There are necessary and sufficient conditions in the literature for a (left) partial order $\le$ on $G$ to extend to a (left) linear order $\le^{\ast}$ on $G$. This shows in particular, that not every partial left order extends to a linear left order in the non-abelian case, even though the group is orderable.   

In the paper "Right-orderability of groups" by Richard Kaye (1998) these conditions are called "a sort of mini completeness/soundness theorem". 
In the paper *"Compactness of the space of left orders"* ([arXiv](https://arxiv.org/abs/math/0606264)) of Dabkowska, Dabkowski, Harizanov, Przytycki and Veve, these conditions are referred to as Conrad's theorem (P. F. Conrad, *Right-ordered groups*, Mich. Math. J. 6(3), 1959, pp. 267–275.) This paper also gives an explicit example of a partial order on the fundamental group of the Klein bottle that does not extend, even though the group is orderable.


Remark: I have edited the answer to include Alexander's useful comments, which helped to clarify the answer.


