Let $H$ be a simple algebraic group of type $\mathbf{G}\_2$ over $\mathbb{C}$.
Let $\rho$ be the standard 7-dimensional complex representation
$$ \rho\colon H=\mathbf{G}\_2\to \mathrm{SO}\_7.$$
We consider the corresponding homomorphism
$$H\to \mathrm{Spin}\_7=G.$$
We regard $H$ as a subgroup of $G$.
Let $T\_H\subset H$ and $T\_G\subset G$ be compatible maximal tori.
We obtain a homomorphism of cocharacter groups
$$\rho\_\*\colon X\_\*(T\_H)\to X\_\*(T\_G).$$
We choose compatible Borel subgroups $B\_H$ and $B\_G$ such that $T\_H\subset B\_H\subset H$
and $T\_G\subset B\_G\subset G$, then we obtain bases consisting of simple coroots in $X\_\*(T\_H)$ and $X\_\*(T\_G)$.
Since we have bases, we can associate with $\rho\_\*$ an integral $3\times 2$ matrix.



> 
> **Question.** How can one compute this matrix?
> 
> 
> 


Note that I need this matrix only modulo 2. 



Matrix from a homomorphism of simply connected groups
Like Jason I'd be more inclined to look at this kind of embedding in a geometric or conceptual way. But for your purposes a concrete description seems needed. For this it might be simpler to work in the Lie algebra setting. I don't see immediately what the group viewpoint does for you, since the centerless group $G\_2$ embeds in either the special orthogonal group or its slightly elusive simply connected covering group Spin. The basic roots-and-weights technology here depends just on the Lie algebra embedding. For this to be made concrete, however, you'd have to line up the two Cartan subalgebras (Lie algebra of maximal tori) in a compatible way and relate the two root systems via simple roots. 


A description of the embedding here is given (in an adapted version going back to one of the Paris seminars) in the first part of section 19.3 of my 1972 Lie algebra textbook. (Needless to say, I haven't spent a lot of time with this material since then.) The main deficiency in the details written down is that a choice of simple roots for $G\_2$ isn't made explicit (though it is in Bourbaki). So you'd have to work that out further. Since the weight lattice and root lattice of $G\_2$ coincide, passage to the duals can be done over $\mathbb{Z}$. 


ADDED: With apologies for the delay in filling in some details, I still have some doubts about the basic setting here. On the level of root lattices it seems fairly clear geometrically. For the embedding $\mathfrak{g}\_2 \hookrightarrow \mathfrak{so}\_7$, one can work over $\mathbb{R}$ (since the Lie algebras are split). Here the bigger root lattice lives in a 3-dimensional euclidean space with standard orthonormal basis $\varepsilon \_1, \varepsilon\_2, \varepsilon\_3$. 


For $B\_3$ there are 9 positive roots. Two simple roots are long: $\alpha\_1 = \varepsilon\_1 - \varepsilon\_2$ and $\alpha\_2 = \varepsilon\_2 - \varepsilon\_3$. A third simple root is short: $\alpha\_3=\varepsilon\_3$. Then the root lattice for $G\_2$ lies in the plane through 0 defined by the condition that coordinates sum to 0. Here you can take a short simple root $\alpha = \alpha\_1$ and a long simple root $\beta = \varepsilon\_2 + \varepsilon\_3 -2\varepsilon\_1$. Then you can compute your $2 \times 3$ or $3 \times 2$ matrix in terms of these bases, keeping in mind that for $G\_2$ the root lattice equals the weight lattice and maps into the root lattice for $B\_3$ (which however has index 2 in its weight lattice).
My quick arithmetic wasn't reliable, but while the matrix you get depends on the choice of bases, the relationship between the two root systems depends only on the Cartan matrices.


