Noise addition. Sum of shot and thermal noise
Wikipedia has an article on this: https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables but when I look at a noise waveform, being random as it is, it seems that there may be some cancellation between the two sources. In terms of amplitude, it looks like they can't add up.


You are correct. The amplitudes don't add up. But the variances (which are proportional to the power in the noise sources) do add up.

So if you have two independent Gaussian random variables with RMS amplitudes \$\sigma_1\$ and \$\sigma_2\$, and we call the RMS amplitude of the sum \$\sigma_N\$, then the variances are \$\sigma_1^2\$, \$\sigma_2^2\$, and \$\sigma_N^2\$, with 

$$\sigma_N^2 = \sigma_1^2 + \sigma_2^2$$

(just like it says in the Wikipedia article)

or

$$\sigma_N = \sqrt{\sigma_1^2 +\sigma_2^2}$$

This is often called "Pythagorean addition" in reference to the Pythagorean theorem. The amplitudes of the two noise sources add up as if they were vectors at right angles to each other.


  Would [the sum] have a Gaussian distribution if I sampled it at 10 MS/s after passing it through a brickwall low pass filter at 1 MHz? (the noise is not white anymore)


It would be Gaussian, but no longer white. Each sample would not be independent of (would have some correlation with) the samples preceding or following it.


  What if I sampled it at just 2 MS/s after the filter? Would the RMS amplitude be the same as in the oversampling case? 


I'm sorry I don't have a confident answer for this. I think the answer is messy because you chose a sampling rate where your filter is just at the Nyquist frequency. If the analog filter is a perfect boxcar filter (it isn't), then you are just on the border between oversampling the noise. If the filter allows any roll-off at all below 1 MHz (it will) then you will see some correlation in the output of your 2 MS/s sampler.