Through my research I have arrived at the following question which I am unable to solve:
Given a triple of 4-tuples, $$(a_1,...,a_4);(b_1,...,b_4);(c_1,...,c_4)$$ (all are vectors, taken from the set of $k$-Hamming weight vectors (vectors with exactly $k$ ones) in $\Bbb{F}_2^n$ for $n>k$) where each of the 4-tuples is a linear dependency (i.e. $a_1+a_2+a_3+a_4=0$, same for $b_i,c_i$'s) match it with a couple of 6-tuples, $$(d_1,...,d_6);(e_1,...,e_6)$$ such that these are also linear dependencies (i.e $d_1+...+d_6=0$, same for $e_i$'s), and $d_i$'s and $e_j$'s are also of $k$ Hamming weight for any $i,j$, and this matching should be one-to-one (Two triples (with order taken for the triples and inside the triples) give distinct two couples).
So far, I have maps for each $t\in\Bbb{F}_2^n$ that match a triple of 4-dependencies to a couple of 6-dependencies, given by: $$f_t((a_1,...,a_4);(b_1,...,b_4);(c_1,...,c_4))=(a_1,a_2,a_3,b_1,b_2,a_4+b_1+b_2);(b_3,c_1,c_2,c_3,t,c_4+b_3+t)$$ but this clearly does not guarantee that the last element in each 6-tuple is of $k$ Hamming weight.
Basically I think that my problem is that I don't know how to use the fact that the 3 dependencies I get as input all have Hamming weight $k$ (That is, that the sum $a_1+a_2+a_3$ also has Hamming weight $k$) to construct the 6 tuples.