Geodesics on zero-curvature regions of closed surfaces of genus > 1 of non-positive curvature Let $M$ be a 2-dimensional Riemannian manifold of non-positive curvature everywhere, of genus > 1.  Let $\textbf{D} \subset \textbf{C}$ be the open unit disc in the complex plane, the universal cover of $M$.  Let $\gamma \subset \textbf{D}$ be a curve representing a geodesic in $M$ which is entirely in a region of zero curvature.  It seems to me, because the definition of geodesic is local, that unless $\gamma$ is tangent to some region $A \subset \textbf{D}$ of negative curvature, $\gamma$ will be a Euclidean line through $\textbf{D}$.  Is this correct?
Secondly, does anybody have any references that I could peruse to learn how a geodesic $\gamma$ which does in fact pass tangent to some $A$ of negative curvature reacts to this region (will it turn into $A$, away from it, etc. and maybe some way of calculating the actual effect)?
Thank you. 
 A: I am answering the question as clarified in your Aug 3 comment. No it is not always possible to identify the universal cover with $D$ so that all zero-curvature geodesics become straight lines.
Begin with a metric in a small region where some 9 zero-curvature geodesics intersect each other but violate the Desargues theorem. For example, begin with a standard Desargues configuration and introduce a bit of negative curvature so that one of the lines misses one of the intersection points. For each of these 9 geodesics, attach a long narrow planar strip to the boundary of the region: each end of the strip is attached to the place where the geodesics meets the boundary, so that the geodesic extended to the strip closes up (and has a neighborhood isometric to a straight cylinder which lies partly in the strip and partly in the original region).
Now we have a surface with boundary, and on this surface some 9 zero-curvature closed geodesics violate the Desargues theorem within a simply connected region. Fill the boundary components by surfaces of sufficiently large genus - so that they can be equipped with a nonpositively curved metric. Now we have a closed surface. In its universal cover, the lifts of the 9 geodesics still violate the Desargues theorem. Hence they cannot be represented by straight lines no matter how you identify the universal cover with $D$.
A: Suppose that $S$ is your Riemannian surface and $X \subset S$ is a flat subsurface (that is, locally isometric to $\mathbb{R}$ with the usual metric).  Let's suppose that $X$ has some nontrivial topology.  For example, $X$ is a unit disk minus one-half of a unit disk, and the core curve of $X$ is essential in $S$. 
You are correct in thinking that the universal cover of $S$ is homeomorphic to $\mathbb{D}$ the unit disk.  However, it is impossible to choose this homeomorphism so that the universal cover of $X$, call it $\bar{X}$, embeds isometrically in $\mathbb{D}$.  You cannot even arrange this up to homothety.  To see this, choose isometric charts for $X$ that lift to give isometric charts for $\bar{X}$.  After choosing where any one chart of $\bar{X}$ goes in $\mathbb{R}^2$ (isometrically!) the positions of all others will be determined.  (This is the so-called "developing map" of $\bar{X}$ and you can think of it as being similar to the process of analytic continuation of a analytic function.) The point here is that the developing map will not be injective - in fact it will have image isometric to $X$ itself. 
I can't think off hand of a reasonable reference (Thurston's book is perhaps an unreasonable reference).  Think about it and ask any local geometers or post more questions on MO. 
