Let $f : X \rightarrow Y$ be a map of topological spaces.
Suppose that $f$ satisfies the following lifting property:
Where the lift is unique up to homotopy of paths. Say a map of topological spaces $f : X \rightarrow Y$ is a formal covering space if it satisfies this lifting property.
If $\Pi_1 : \text{Top} \rightarrow \text{Grpd}$ from topological spaces to groupoids is the fundamental groupoid functor, then we have the following unique lifting property:
Say a groupoid covering is a map of groupoids that satisfies this lifting property. The category of covering groupoids of $\pi_1(X)$ is equivalent to the category of $\pi_1(X)$-sets for a groupoid $X$.
My questions are:
1) Is there a correspondence theorem for the formal covering spaces?
2) Are formal covering spaces local homeomorphisms?
If you understand a homotopy of paths as usual as a homotopy keeping the end-points fixed, then your lifting condition implies that all fibers of $f$ are singletons. In fact, let $y \in Y$ and $c$ be the constant path at $x$. It must have a lift which is unique up to homotopy of paths which would be impossible if the fiber over $y$ had more than one point. Thus $f$ is a continuous bijection. Such maps are local homeomorphisms iff they are homeomorphisms.
Now let $Y = \{0\} \cup \{1/n \mid n \in \mathbb N\}$ with the subspace topology inherited from $\mathbb R$ and $X = \mathbb N$ with the discrete topology. Define $f : X \to Y, f(1) = 0, f(n) = 1/(n-1)$ for $n > 1$. This is a continuous bijection, but no homeomorphism. It has your lifting property because all paths in $Y$ are constant.
Note that the situation does not really change if you only consider liftings $\tilde u$ of $u : I \to Y$ with a prescribed $\tilde u (0) \in f^{-1}(u(0))$. In that case the fibers of $f$ can only have constant paths. The above example remains valid.
I doubt that this is what you want.