It's true it can be quite difficult to use trial and error to find orthogonal mates for Latin square. However, that doesn't mean there's a better method.
It can even be impossible to find a mate: bachelor Latin squares (those without orthogonal mates) exist for all orders except $1$ and $3$. For example, the Cayley table of $\mathbb{Z}_{2n}$ has no orthogonal mate, for all $n \geq 1$.
I suspect there is some misinterpretation as to what the lecturer was talking about here.
One relationship between orthogonal Latin squares and magic squares is that if $L$ and $M$ are orthogonal Latin squares of order $n$, then $nL+M$ is a magic square.