In order to make sense of your question one has to specify what do "space" and "cover" mean. Without further adjectives, they usually mean "a topological space" and "a topological covering map." However, it is clear from your comments, this is not what you are actually interested in. One can define the notion of a covering in other categories of "spaces," for instance, for triangulated manifolds, complex manifolds, and for Riemannian manifolds. Let me first discuss the case of Riemannian manifolds. I will assume that you know at least basic Riemannian geometry. Given a connected Riemannian manifold $(M,g)$ (where $g$ is the Riemannian metric tensor on $M$, aka a Riemannian metric), and a local diffeomorphism
$$
f: N\to M
$$
one defines the pull-back Riemannian metric $h=f^*(g)$ on $N$ by the rule
$$
h(u,v)=g(Df_p(u), Df_p(v)), p\in N, u, v\in T_xM.
$$
When we equip $N$ with the Riemannian metric $h$, the map
$$
f: (N, h)\to (M, g)
$$
becomes a (Riemannian) isometric map. This does not mean, of course, that $f$ preserves the Riemannian distance function on $N$, only that $Df$ preserves the inner product on tangent spaces. Now, suppose that $(M,g)$ is a compact Riemannian surface (which is not the same as a Riemann surface). Let $f: \tilde M\to M$ denote the universal covering map. One can then pull-back the Riemannian metric $g$ to $\tilde M$, $h:= f^*(g)$. One then can ask for conditions under which $(\tilde M, h)$ is isometric to the Euclidean plane (equipped with the standard flat Riemannian metric $dx^2 + dy^2$). An isometry in the sense of Riemannian geometry in this case is the same as an isometry in the sense of metric geometry, i.e. it will preserve the Riemannian distance function. (Differential geometers frequently regard isometric Riemannian manifolds as "the same". I will adopt this viewpoint.) It turns out that there are several equivalent conditions for $(\tilde M, h)$ to be isometric to the Euclidean plane (with the standard flat metric),
one of which is that the Gaussian curvature of the metric $g$ is identically zero (i.e. the metric $g$ is flat). Gauss-Bonnet formula then implies that $\chi(M)=0$, i.e. that $M$ is diffeomorphic to the torus or to the Klein bottle. Conversely, both torus and the Klein bottle admit flat Riemannian metrics. Thus, one obtains:
Theorem. A compact connected surface $M$ admits a Riemannian metric $g$ such that the universal cover of $(M,g)$ (equipped with the pull-back Riemannian metric) is isometric to $(\mathbb R^2, dx^2 + dy^2)$ if and only if $M$ is diffeomorphic (equivalently, homeomorphic) to the torus or the Klein bottle. Moreover, if $(M,g)$ is a flat compact connected surface, then the universal cover of $M$ (equipped with the pull-back Riemannian metric) is isometric to
$(\mathbb R^2, dx^2 + dy^2)$.
Is it enough for your purposes? Maybe. Or maybe not. It might be the case that you are actually not dealing with smooth surfaces and instead you are dealing with triangulated surfaces $M$. Accordingly, instead of a Riemannian metric on such a surface, you are dealing with a piecewise-Euclidean metric, i.e. a metric (understood in the sense of metric geometry, i.e. a distance function) on each 2-dimensional simplex of the triangulation, such that each simplex is isometric to a simplex in the Euclidean plane (equipped with the "standard" distance function). Furthermore, whenever two 2-dimensional simplices $\Delta_1, \Delta_2$ share an edge $e$, the length of $e$ computed in terms of the metric on $\Delta_1$ is the same as the length of $e$ computed in terms of the metric on $\Delta_2$. Such substitute for a Riemannian metric appears frequently in applied differential geometry and data science. Given the above data (a triangulation together with a metric on each simplex) one defines a path-metric $d$ on the surface $M$, computed by minimizing the arc-length of piecewise-linear paths between the given pair of points in $M$. Furthermore, given the universal covering $f: \tilde M\to M$, one can pull-back the triangulation from $M$ to $\tilde M$ and, accordingly, equip each 2-dimensional simplex $\tilde \Delta$ in $\tilde M$ with a metric such that the restriction map
$$
f: \tilde \Delta\to \Delta=f(\tilde \Delta)\subset M
$$
is an isometry. (This is an analogue of taking the pull-back of a Riemannian metric.) Now, one can ask various questions about the geometry of $\tilde M$ equipped with this structure. For instance, one can consider $\tilde d$, the path-metric on $\tilde M$ defined as before. The map
$$
f: (\tilde M, \tilde d)\to (M,d)
$$
is not an isometry in the sense of metric geometry. Nevertheless, it preserves lengths of paths and restricts to an isometry on each 2-dimensional simplex. One can ask for conditions under which $(\tilde M, \tilde d)$ (as a metric space) is isometric to the Euclidean plane with the standard distance function. The answer will be pretty much the same as in the Riemannian setting: If and only if the combinatorial curvature of $M$ is zero, i.e. the total angle around each vertex of the triangulation is $2\pi$. It will imply, again, that $M$ is either homeomorphic to the torus or the Klein bottle. In practice, zero curvature is too much to ask and one can ask for some softer conditions. For instance, one can consider the graph in $\tilde M$ given by its triangulation. (I will assume that this graph is infinite, equivalently, that $\tilde M$ is noncompact.) One defines the random walk on this graph, for instance, by declaring the transition probability to be the same for all edges. Then one can ask for the random walk to be recurrent, i.e., that with probability 1 it returns to every given compact subset infinitely often. This, turns out again imply that $M$ is homeomorphic to the torus or the Klein bottle. Moreover, while $\tilde M$ in our setting does not have a natural Riemannian metric (because the total angles at vertices need not be $2\pi$), it has a canonical complex structure and one can ask if this Riemann surface is conformally isomorphic (biholomorphic) to the complex plane. This again turns out to be equivalent to the assumption that the random walk is recurrent, and, equivalently, that $M$ is homeomorphic to the torus or the Klein bottle.
I stop here since my answer is already too long...
