I remember back in school (some time ago) we were taught to solve problems such as the following:
If $6$ men can do $1/3$ of work in $10$ days then how many days would it take $4$ men to do $2/3$ work?
Our teacher constructed the following diagram to solve the problem
and solved it as follows:
$$6\times 10\times \frac 23 = 4\times x\times \frac 13 \implies x=\frac 13$$
Now I am already familiar with the scenario on when to use cross-multiplication
such as (when both the columns in the next row increase or decrease simultaneously)
and I am also familiar when to use horizontal multiplication (When one column in next row increases and the other decreases or vice versa such as no of men and days example)
Coming back to the main question is there a key to remembering which column gets placed in the center or any other points to consider when constructing such a diagram ?
I don't think there's much point in learning such ad-hoc tricks to solve particular forms of problems. Rather, one should understand how to translate the problem into algebra.
The basic idea is that the amount of work done is assumed to be proportional to both the number of workers and the time taken: thus $w = c m t$ for some constant $c$, where $w$ is the amount of work, $m$ the number of workers and $t$ the time. This is equivalent to $c = m t/w$. So if we have one data point with $m = 6$, $t =10$ and $w = 1/3$, and our new scenario involves $m=4$, $t=x$ and $w=2/3$, that says $\dfrac{6 \times 10}{1/3} = \dfrac{4 \times x}{2/3}$, which you then solve for $x$.