At first: sorry, I could't find the source you were looking for, but I'm so free to sketch two ways how you in my opinion get very fast to the informations you need (especially the latter, generic, might be the reason rarely somebody wants to work out all relations?)

Maybe you've a more specific question than just knowing the entire table, which schould be quite huge?

By the way: With Chevalley-group you mean the finite simple group? But not the twisted one? I assume yes on both, as this is what you wrote ;-)

**1)** The **probably easiest way** to directly and elementary get ahold of $G_2$ is as **folding** of $D_4$ by the cyclic interchange of the three outer nodes. I.e. take $D_4=so_8$ and take the sub-algebra, sub-root-system, etc. fixed by the outer order-3-automorphism often named **triality**. I'm not 100% sure and don't have a proper source, where this is done for Chevalley groups, but I'm pretty positive, you just do it ;-)

....do **NOT confuse** this with the construction of the Steinberg group $^3D_4(q^3)$, where one takes (analogously to the construction of unitary groups) the elements, where the action of the automorphisms matches the order 3 galois field automorphisms forced to exist by taking $\mathbb{F}_{q^3}$.

**This way you're commutator relations of $G_2(q)$ are just a subset of those in $SO_8(\mathbb{F}_q)$, where calculating is obvious (but tedious?)**

**2)** The **more uniform, direct way** would be starting with a **Chevalley basis** of the Lie algebra $G_2$ (which I'm sure is not hard and/or can be found in literature - I'll look into it, if I find time...) - the relations are then just exponentiated to the Chevalley group.

**This way you can read off the commutator relations just from the structure constants of the Lie algebra $G_2$, that I'm sure are worked out at many places**