The autocorrelation signals you have shown are biased autocorrelation. The issue is that the higher lags have fewer data points to be used to estimate the correlation at those lags.
Another issue is that your original signal appears to have a DC (constant) offset. When the biased autocorrelation estimator is applied to a signal with a constant offset, the result is a triangular shaped function.
Detrend. de-mean, or apply a DC blocker to your signal before taking the autocorrelation and you'll get something more sensible (with fewer artifacts from known causes).
Here's an update.
The plots you have are not very useful because of the two issues I raised above: the mean is non-zero (hence the triangular baselines) and the biased estimator is used (hence the triangular envelopes).
Let's look at a simple example.
First, let's take a sinusoid offset by a constant value.
and find its autocorrelation.
Because of the durations I've chosen, the two effects I mentioned are more apparent: the "baseline" (value about which the autocorrelation swings) is triangular in shape and the "envelope" (the outline) is also triangular.
First, let's remove the mean of the signal before taking the autocorrelation.
That makes the baseline zero, so the only effect left is the envelope.
Now let's try to correct for the envelope.
That matches a little better with the understanding that the autocorrelation of $\sin(\omega t + \phi)$ should be $\cos(\omega t)$, though it's not perfect because of the shortcut I've taken to form it (see code below).
Now, back to the actual question you're asking:
How to interpret values of the autocorrelation sequence?
As Knut says, it says given the value of the signal at time $n$, what is the signal likely to be at time $n+m$?
For signals like sinusoids, this means that the autocorrelation value at $n+m$ value will depend on the frequency of the sinusoid.
For signals like (band-limited) white noise, the autocorrelation value will be zero (or likely close to it for any particular noise realization).
import matplotlib.pyplot as plt
import numpy as np
T = 1000
fs = 44100
t = np.arange(T)/fs
omega = 2*np.pi*1000
phi = 2*np.pi*0.8978941234
x = 1 + np.sin(omega*t + phi)
Rxx = np.correlate(x,x, mode='full')
Rxx2 = np.correlate(x-np.mean(x), x-np.mean(x), mode='full')
Rxx2un = np.divide(Rxx2, np.bartlett(len(Rxx2)))