How do EM waves travel infinite distances if it’s source has negligible effect over the infinite distance? I understand EM waves as not something that travels, rather an oscillating disturbance in the EM field, which is caused due to an accelerating electrical charge, around which the field updates at the speed of light, but over large distances, it takes time for it to update, and if the electrical charge vibrates, then the EM field again has to update accordingly, forming a wave-like disturbance which keeps “travelling” or updating at the speed of light over distance in the EM field.

My problem is that if an electrical charge has an EM field around it, and it accelerates, then the EM field updates around it, but if the electrical charge has negligible effect say some distance away from it, then when the EM field updates after the charge moves, that part should still have negligible effect and also when it comes back to it’s original position and vibrates, then the EM wave should only travel in the electric field of the charge, not outside it as that area has an almost zero value to it, which should not change as it accelerates and vibrates.

OR A BETTER EXPLANATION

My problem is that when an electric charge accelerates, and the EM field updates, and the charge comes back to it’s original position(it’s vibrating), then the electric charge should have negligible effect on the EM field, say some distance away from the charge, and when it updates and forms a wave-like disturbance, it should not travel outside the electric field as it has an almost zero value some distance away from the charge, which should not update to a different value, as when the electric charge accelerated and moved, it still had an almost zero value at that location.

So how do EM waves travel infinite distances in vacuum?
 A: 
if the electrical charge has negligible effect say some distance away from it, then when the EM field updates after the charge moves, that part should still have negligible effect and also when it comes back to it’s original position and vibrates, then the EM wave should only travel in the electric field of the charge, not outside it as that area has an almost zero value to it, which should not change as it accelerates and vibrates

This is actually not correct. If you look at the Lienard Wiechert fields you see that the E field has two terms: $$ \mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \left(\frac{q(\mathbf{n}_s - \boldsymbol{\beta}_s)}{\gamma^2 (1 - \mathbf{n}_s \cdot \boldsymbol{\beta}_s)^3 |\mathbf{r} - \mathbf{r}_s|^2} + \frac{q \mathbf{n}_s \times \big((\mathbf{n}_s - \boldsymbol{\beta}_s) \times \dot{\boldsymbol{\beta}_s}\big)}{c(1 - \mathbf{n}_s \cdot \boldsymbol{\beta}_s)^3 |\mathbf{r} - \mathbf{r}_s|} \right)_{t_r}$$ The first term is the "Coulomb" field that you are describing above. Note that it falls off as $1/|\mathbf{r} - \mathbf{r}_s|^2$. So the Coulomb portion of the field decays pretty rapidly.
In contrast, note the second term. This term is proportional to the acceleration, $\dot{\boldsymbol{\beta}_s}$, and it falls off as $1/|\mathbf{r} - \mathbf{r}_s|$, which is much more slowly than the Coulomb field. This is the radiation term. Because it falls off so much more slowly than the Coulomb field, the radiation field can be large even when you are in an area where the Coulomb field is negligible.
Indeed, if you are in a location where $1/|\mathbf{r} - \mathbf{r}_s|$ is significant and $1/|\mathbf{r} - \mathbf{r}_s|^2$ is negligible then you will start with negligible field from the stationary charge, get a significant field from the accelerating charge, and then go back to negligible field once the charge stops accelerating. The Coulomb field decays at a different rate than the radiation field.
A: Besides the point of view that electromagnetic radiation is a disturbance in an EM field (developed for the consideration of atomic processes), there is another one. This was previously developed by Einstein and states that EM radiation consists of quanta created by the relaxation of excited subatomic particles. These quanta, later called photons, are indivisible units that move forward in matter-free space until they are absorbed again by a particle.
You can learn about the nature of photons by periodically setting electrons in motion on a rod. Then photons are emitted that are polarised. This means that you can measure an oscillating electric field parallel to the rod and an oscillating magnetic field perpendicular to the rod. In a vacuum, these two fields are perpendicular to each other (and perpendicular to the direction of propagation).
For processes describing the interactions in the atom, quantum mechanics and the related descriptions QED etc. are successful. For processes that you describe and that involve the unhindered propagation of EM radiation, the description with photons is the more comprehensible one. The following applies: Once the photons have left the near field of the emitter, they are no longer subject to any influence. Except for gravitation. But that is another story.
