The problem is that the conditional constraint is not convex: take
$z_1=(x_1,x_3,x_6)=(1,0,1)$ and $z_2=(1,1,0)$, then
$$
\frac12(z_1+z_2)=(1,1/2,1/2)
$$
is not in the set. Therefore, it cannot be, in general, described by linear inequalities which are convex (unless the rest of the constraints are so special that the intersection takes away this trouble, but it is very particular case).
What you can do is to split your LP in two optimizations: one with $x_6>0$ and $x_1+x_2=1$ and another with $x_6=0$. Take the solution that has the smaller objective.
UPDATE: As I understand, the problem takes the following form: denote the variables in the first set as $(x_1,x_2,x_3,x_4)$ and those in the second one as $(y_1,y_2,y_3,y_4)$, so the overall vector looks like
$$
(x_1,y_1,x_2,y_2,x_3,y_3,x_4,y_4),\qquad x_i,y_j\in[0,1]
$$
and no more constraints except for the mentioned in the question that
$$
k=1,2,3\colon\quad s_k=\sum_{j=1}^ky_j>0\quad\Rightarrow\quad t_k=\sum_{i=k+1}^4x_i=0
$$
which can be shortly written as $s_kt_k=0$. Thus, the problem is
- Some linear equalities and inequalities, and
- Some pairwise products are zeros.
This is precisely the form of the KKT condition for linear programming (feasibility and complementary slackness principle). The problem is not linear, but can quite effectively (in polynomial time) be solved by the primal-dual interior point method for LP. See, for example, Chapter 14 in this book.