(Almost) double light speed

One of the results of special relativity is that a particle moving at the speed of light does not experience time, and thus is unable to make any measurements. In particular, it cannot measure the velocity of another particle passing it. So, strictly speaking, your question is undefined. Particle #1 does not have a "point of view," so to speak. (More precisely: it does not have a rest frame because there is no Lorentz transformation that puts particle #1 at rest, so it makes no sense to talk about the speed it would measure in its rest frame.)

But suppose you had a different situation, where each particle was moving at $0.9999c$ instead, so that that issue I mentioned isn't a problem. Another result of special relativity is that the relative velocity between two particles is not just given by the difference between their two velocities. Instead, the formula (in one dimension) is

$$v_\text{rel} = \frac{v_1 - v_2}{1 - \frac{v_1v_2}{c^2}}$$

If you plug in $v_1 = 0.9999c$ and $v_2 = -0.9999c$, you get

$$v_\text{rel} = \frac{1.9998c}{1 + 0.9998} = 0.99999999c$$

which is still less than the speed of light.

This is what special relativity is all about..

In special relativity you cannot simply state that particle 2 is moving at c+c=2c in a reference frame where particle 1 is at rest.

Speeds add like this (easily found in wikipedia):

$$v_2^{'} = \frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$$

i.e. the speed of particle 2 $v_2'$ in a reference frame where particle 1 is at rest is

$$v_2^{'} = \frac{c+c}{1+1} = c$$

you cannot move faster than at the speed of light in the vacuum.

Easy. you cannot sit on a particle moving at the speed of light. If you could then you would be massless, and unable to sum properly

In any case, there is no reference-frame moving with either photon, so no operational way to measure relative velocities between them. relative velocities have meaning only from a inertial frame. There are no inertial frames moving with the photon, otherwise this frame would measure that photon to be at rest