First, as has been mentioned, this is a directed acyclic graph. It sounds like your "linear sequence form" is an arrangement of the vertices in a left-to-right sequence $\langle v_1, v_2, \dots , v_n\rangle$ such that any edge goes from left to right in the sequence. If so, you're asking for what's commonly known as a topological sort of the vertices.
For your graphs an easy way to do this is to order the vertices first by weight, where the weight of a vertex $w(i, j)=i+j$ and then within each weight order the vertices by their first coordinate. For example, if $n=2$ we'll have the linear order
(0,0), (0, 1), (1, 0), (0, 2), (1, 1), (2, 0), (1, 2), (2, 1), (2, 2)
Now look at the edges. From $(i, j)$ we'll have edges to $(i+1, j), (i, j+1), (i+1, j+1)$, as long as $i+1\le n$ and $j+1\le n$. Each of these edges will have weight strictly greater than $i+j$ so each of the edges from $(i, j)$ will end at a vertex of greater weight, hence to the right of $(i, j)$. This also shows that the graph is acyclic.
Notice that there are other ways of arranging the vertices in a line (simply permute the edges of equal weights).