To expand on what Alexander said, we aren't looking at group actions of $\pi_1(X)$ on $\widetilde{X}$, but we're looking at the group actions that are compatible with the covering map. Let's go over the construction to see what is happening.
If we have a covering map $p:E\to B$ of spaces, where $b\in B$ is a fixed basepoint, then for any path $\gamma:[0,1] \to B$, with $\gamma(0)=\gamma(1)=b$, $x\in p^{-1}(b)$, there is a unique lift $\gamma_x:[0,1]\to E$ of $\gamma$ with $\gamma_x(0)=x$. Note that by a lift of $\gamma$, we mean that $p(\gamma_x(t))=\gamma(t)$. Let us assume that $\gamma$ is a loop, that is $\gamma(0)=\gamma(1)=b$,
We assert the following facts:
(1) The map $x\mapsto \gamma_x(1)$ is in fact an automorphism of $p^{-1}(b)$.
(2) This map depends only on the class of $\gamma$ in $\pi_1(B)$.
(3) This map $\pi_1(B) \to \operatorname{Aut}(p^{-1}(b))$ is actually a group homomorphism.
Given an automorphism of $\phi$ of $p^{-1}(b)$, there is at most one automorphism $\psi$ of $E$ which both restricts to $\phi$ and satisfies $p(\psi(e))=p(e)$ for all $e\in E$. Of course, not every automorphism lifts. For example, if $\mathbb{R}\to S^1$ is the standard covering map, the automorphism of $\mathbb{Z}$ (viewed as a space, not a group) defined by $z\mapsto (-1)^{z} z$ doesn't extend to a continuous automorphism of $\mathbb{R}$. However, by using the path lifting property, we can show that all the automorphisms we've generated actually do extend (prove this!).
So what we have shown is that there is a unique action of $\pi_1(B)$ on $E$ which is compatible with the projection map $p$, in the sense that the action comes from path lifting. There are probably other characterizations of the action in terms of an action of $\pi_1(B)$ on $\pi_1(E)$.
Of course, there are lots of other actions. For example, given any homomorphism from $\pi_1(B)$ to itself we can compose these with our given action to get a new one
$\pi_1(B) \to \pi_1(B) \to \operatorname{Aut}(E)$
For example, composing with the trivial group homomorphism, we get the trivial action (all loops act trivially). That doesn't take away from the fact that there is a canonical action.
In the case of the universal covering space, there is a very simple description of the action. One of the constructions of $\widetilde{X}$ is as the space of homotopy classes (preserving the endpoints) of paths in $X$ which start at the basepoint. The covering map sends a path to its endpoint. The action of $\pi_1(X)$ takes a loop $\gamma$ and a path $\rho$ and gives the path obtained by following $\gamma$ and then following $\rho$.
The point is that all of this fits together very nicely: For every covering of a space $X$, we have a canonical action of $\pi_1(X)$, and if we have a map of two covering spaces of $X$ (a continuous map which is compatible with the projection maps), then this map is compatible with the action of $\pi_1(X)$. If $X$ is nice (so that the universal exists), and if we look only at connected covering spaces, then there is always a unique map of covering space from the universal covering to any other given cover. This map is the unique map which has the property you stated in your question.
To go back to the original question, we have a natural action of $\mathbb{Z}^2$ on $\mathbb{R}^2$. Namely, we view both as groups under addition, and the action comes from the natural inclusion.
Every subgroup of $\mathbb{Z}^2$ is either isomorphic to $\{0\}$, $\mathbb{Z}$, or $\mathbb{Z}^2$, although there are many different subgroups of the latter two types, which up to automorphism correspond to orbits of points or pairs of points in $\mathbb{Z}^2$ under the action of $\operatorname{SL}_2(\mathbb{Z})$. While the quotient of any two different subgroups gives a different covering map, we have that (in this special case, with these particular actions) the quotient of $\mathbb{R}^2$, up to homeomorphism, by the subgroup depends only on the isomorphism class of the subgroup. For isomorphic but non-equivalent subgroups, even though the space on top is the same, the action of $\mathbb{Z}^2$ is different. So in fact, this gives an example of how a group can act on a space in multiple ways.
But what if we don't want to use this action? What spaces can you get? Well, then depends on what kind of action you want to take. If you act by $(m,n).(x,y)=(-1)^m x, (-1)^n y)$, you will something that looks like the first quadrant of the plane. If you act by $\mathbb{Z}$ with $n.(a,b)=(2^n a,2^n b)$, the quotient is almost the torus: every point other than the original can be moved inside the the annulus between the unit circle and the circle of radius $2$, and so the quotient of the action on $\mathbb{R}^2\setminus \{0\}$ is the torus, but then you have an extra point. You can get all sorts of interesting things if you are okay with the action not treating every point equally. I am not sure what the classification of such quotients looks like, or even what familiar things you can get from such quotients.
For a nice action, where each point has the same stabilizer, and modulo this stabalizer the action is free, you will get some surface whose fundamental group is a quotient of $\mathbb{Z}^2$, and whose universal cover is $\mathbb{R}^2$. The first condition rules out getting things like higher genus surfaces and the Klein bottle, whose fundamental groups are not abelian.. The second condition rules out the projective plane, whose universal cover is $S^2$.
Update in response to the clarification request:
When you have a group acting on a space, quotienting out by a subgroup will generally depend on the specific subgroup, and not just the isomorphism type of the subgroup. This is easy to see for non-covering-space examples, where the group doesn't act in a uniform way: consider $\mathbb{Z}^2$ acting on $\mathbb{R}$ where the second copy of $\mathbb{Z}$ acts trivially. Quotienting out by one copy of $\mathbb{Z}$ does nothing, and by the other gives you the circle.
However, covering maps are nice (the action of the fundamental group of the base is uniform). Let's just consider the case of the universal cover.
Suppose that $H_1,H_2\subset G=\pi_1(X,x)$, are subgroups of the fundamental group which are abstractly isomorphic, but which are not mapped into each other by any automorphism of $G$. For example, we could have $4\mathbb{Z}\subset 2\mathbb{Z} \subset \mathbb{Z}$. The bundles $\widetilde{X}/H_i \to X$, $i=1,2$ will usually be non-isomorphic. For example, if $H_i$ has finite index, then $\widetilde{X}/H_i \to X$ will be a finite covering of degree equal to that index. For the example $4\mathbb{Z}\subset 2\mathbb{Z} \subset \mathbb{Z}$, the two subgroups have index $2$ and $4$.
However, what if we ignore the bundle structure? I believe that $\widetilde{X}/H$ spaces need not be determined by the isomorphism type of $H$. Unfortunately, in one or two dimensions, we can't get any interesting counterexamples: $\widetilde{X}/H$ has fundamental group isomorphic go $H$, and surfaces are determined by their fundamental groups. Additionally, up to homotopy, there are no counterexamples if $\widetilde{X}$ is contractible, as we would have an Eilenberg-Maclane space, which is up to homotopy determined by its fundamental group. See this wikipedia article for more information on EM-spaces.
So is there a conterexample? Probably, but I do not know it. It will require at least $3$ dimensions and a covering space which is not contactable. Still, I have no good a priori reason to believe that quotienting the universal cover by a subgroup depends only on the isomorphism type of the subgroup, and you shouldn't bu lulled into believing this is true in general until you see a theorem asserting it. I don't even know where to begin constructing a counterexample, as the obvious algebraic invariant that might help, the quotient $G/H$, is lost when you forget the bundle structure.
Best Answer
Here is why the plane $\mathbb R^2$ is not a covering of $\mathbb {RP}^2$, and thus even less a universal covering. Consider the univeral covering map $$p:S^2\to \mathbb {RP}^2:(x,y,z)\mapsto [x,y,z]=(x,y,z)/\pm Id$$ If there existed a covering map $f:\mathbb R^2\to \mathbb {RP}^2$, by simple connectedness of $S^2$ the map $p$ could be lifted to a covering map $P:S^2\to \mathbb R^2$ satisfying $p=f\circ P$.
But then the image $P(S^2)\subset \mathbb R^2$ would be bounded by compacity of $S^2$ and thus $P$ would certainly not be surjective.
This however is a contradiction: the covering map $P:S^2\to \mathbb R^2$ must, like all covering maps with connected codomain, be surjective.