Over $\mathbb Q$ the three elements $1+X_1, 1+X_2, 1+X_1+X_2$ generate a free subgroup of rank 3. Indeed, let $G$ be the group generated by these three elements. Consider the $2\times 2$-integer matrices $A=1+2E_{1,2}$, $B=1+2E_{2,1}$, where $E_{i,j}$ is the matrix with 1 in the $i,j$-place and 0 everywhere else. It is well known (Poincare and Sanov) that $A,B$ generate a free group of rank 2. Now consider three matrices $X=A^2B^2, Y=A^3B^{-3}, Z=A^5B^{-1}$. It is easy to check that $X+Y-1=Z$. Hence there exists a homomorphism from $G$ to the group generated by $X,Y,Z$ (it follows from the fact that there exists a ring homomorphism from the free non-commutative ring Ring$\langle X_1, X_2\rangle$ to the ring of matrices generated by $X-1, Y-1$ and under this homomorphism $1+X_1$ is mapped to $X$, $1+X_2$ is mapped to $Y$, $1+X_1+X_2$ is mapped to $Z$). But the subgroup of the free group Group$\langle A,B\rangle$ generated by $A^2B^2, A^3B^{-3}, A^5B^{-1}$ is free of rank 3 (apply Stallings foldings). Hence the group $G$ is also free of rank 3. In order to do it over the field ${\mathbb F}_2={\mathbb Z}/2{\mathbb Z}$, you need to consider matrices over the ring of polynomials ${\mathbb F}_2[t]$. The problem about arbitrary linear forms is more complicated but I think it can be treated the same way.
Edit: As Andreas correctly pointed out, there is no homomorphism from the ring of the formal power series to the ring of matrices over $\mathbb Q$. The argument above only shows that the semigroup generated by these elements is free.
