Let $X$ and $G$ be locally compact, Hausdorff and second-countable spaces such that $G$ is an Abelian topological group with respect to that topology. Suppose that we have a continuous, free, and proper action of $G$ on $X$. That is, a continuous map $X\times G \to X$, $(x,g)\mapsto x+g$ such that for all $x\in X$ we have $\{g\in G: g+x=x\}=\{0_G\}$. Let $\pi:X\to X/G$ be the quotient map (with $X/G$ endowed with the quotient topology).
Is it true that this bundle map is locally trivial? That is: Is there a covering of $X/G$ by open sets $\{U_i\}_{i\in I}$ such that for every $i\in I$ there is a homeomorphism $\phi_i:\pi^{-1}(U_i)\to U_i\times G$ that is equivariant (i.e., that $\phi_i(x+g)=\phi_i(x)+g$ where addition on $U_i\times G$ is understood in the $G$ coordinate)?
There is a result of Gleason https://www.ams.org/journals/proc/1950-001-01/S0002-9939-1950-0033830-7/S0002-9939-1950-0033830-7.pdf that implies this if $G$ is a compact Lie group. But can we drop those assumptions to those that I wrote before? (maybe we need the Lie property?)
The version you propose does not work. The question whether we can drop the Lie property seems interesting, but also too difficult to me, so I will take the cheap way out and illustrate that we cannot drop the compactness property (that said, I believe the result will be true again once you add the assumption that the action should be proper).
Let $X=S^1$ and $G=\mathbb{Z}$. Choose an irrational $r\in[0,1]$ and let $\mathbb{Z}$ act on $S^1$ via $n.z=e^{2\pi nr}z$. Certainly, $S^1$ and $\mathbb{Z}$ are second-countable LCH spaces ($S^1$ is compact, even) and $\mathbb{Z}$ is an abelian topological group. The action of $\mathbb{Z}$ on $S^1$ is free (exercise; this is where it matters that $r$ is irrational) and clearly continuous ($S^1$ is itself a topological group and this is just the left translation action of $S^1$ on itself restricted to the cyclic subgroup generated by $e^{2\pi r}\in S^1$). The quotient $S^1/\mathbb{Z}$ is very ugly, e.g. it is not even $T_1$. I claim that $\pi\colon S^1\rightarrow S^1/\mathbb{Z}$ is not a covering space, which contradicts your claim. Assume to the contrary that it is a covering space. Now, every covering space trivializes over some non-empty closed subspace of the base (exercise), so there is a closed $\emptyset\ne X\subseteq S^1/\mathbb{Z}$, such that $\pi^{-1}(X)\cong X\times\mathbb{Z}$. However, $\pi^{-1}(X)\subseteq S^1$ is closed and $S^1$ is compact, whence $\pi^{-1}(X)$ is compact, but $\pi^{-1}(X)\cong X\times\mathbb{Z}$ and $X\ne\emptyset$ then implies that $\mathbb{Z}$ is compact, which is absurd.