A property of a linear image of the cube

The answer to your question is NO, and here is a counterexample.

Let $L(x,y,z)=(x-z,y-z)$, $w_1=(1,1)$, $w_2=(-\frac{1}{2},0)$, $v_1=(x_1,y_1,z_1)$, $v_2=(x_2,y_2,z_2)$.

Clearly $L(v_1)=w_1$ iff $v_1$ is of the form $v_1=F_1(z_1)=(1+z_1,1+z_1,z_1)$, and this $F_1(z_1)$ is in $[0,1]^3$ iff $z_1=0$. In particular, $w_1\in H$.

Clearly $L(v_2)=w_2$ iff $v_2$ is of the form $v_2=F_2(z_2)=(z_2-\frac{1}{2},z_2,z_2)$, and this $F_2(z_2)$ is in $[0,1]^3$ iff $z_2\in[\frac{1}{2},1]$. In particular, $w_2\in H$.

Moreover, the vector $F_3(z_1,z_2)=F_1(z_1)+F_2(z_2)=(\frac{1}{2}+(z_1+z_2),1+z_1+z_2,z_1+z_2)$ is in $[0,1]^3$ iff $z_1+z_2=0$. In particular, $w_1+w_2\in H$ (take $z_1=z_2=0$).

Finally, if we want all three of $F_1,F_2,F_3$ to live in $H$, we get the impossible system $z_1=0,z_2\in[\frac{1}{2},1],z_1+z_2=0$. This finishes the proof.