Let $\mu(n)$ denote the Mobius function, let $\lambda(n)$ denote the Liouville function, and let $M(x) = \sum_{n \leq x} \mu(n)$ and $L(x) = \sum_{n \leq x} \lambda(n)$ denote their summatory functions. The Riemann hypothesis implies $M(x) = O_{\varepsilon}(x^{1/2 + \varepsilon})$ and $L(x) = O_{\varepsilon}(x^{1/2 + \varepsilon})$.
In fact, if one assumes RH and the simplicity of the zeroes of $\zeta(s)$, then one can show that
\[\frac{M(x)}{\sqrt{x}} = \sum_{\rho} \frac{1}{\zeta'(\rho)} \frac{x^{i\gamma}}{\rho}
\]
for $x$ a positive real that is not an integer. Here the sum is over the nontrivial zeroes $\rho = 1/2 + i\gamma$ of $\zeta(s)$.
Similarly, one can show under the same hypotheses that
\[\frac{L(x)}{\sqrt{x}} = \frac{1}{\zeta(1/2)} + \sum_{\rho} \frac{\zeta(2\rho)}{\zeta'(\rho)} \frac{x^{i\gamma}}{\rho}
\]
This explains why the functions look similar: as $x^{i\gamma} = \cos(\gamma \log x) + i \sin(\gamma \log x)$, these functions basically look like sums of trigonometric waves of decreasing amplitude and frequency.
So if one plots the difference, one finds that
\[\frac{L(x) - M(x)}{\sqrt{x}} = \frac{1}{\zeta(1/2)} + \sum_{\rho} \frac{(\zeta(2\rho) - 1)}{\zeta'(\rho)} \frac{x^{i\gamma}}{\rho}.
\]
What you are seeing in your plots is the presence of the "main term" $1/\zeta(1/2)$. However, the contribution of the trigonometric waves cannot be ignored: although usually the "secondary terms" are not very large, very occasionally they can be big (and if one assumes a conjecture called the linear independence hypothesis, one can show that the secondary terms can get arbitrarily large).
For more on this phenomenon, see e.g. my paper here:
http://arxiv.org/abs/1108.1524