How to Form a Row-Complete Latin Square This is one of the descriptions I've seen online:

For any even $n$, say $n = 2m$, a row complete Latin square of order $n$ can be formed by writing down

$$0, 1, 2m - 1, 2, 2m - 2, 3,\ldots, m + 1, m$$

as the first row and then developing subsequent rows by adding $1$ modulo $n$.

I'm not quite clear on how that goes, such as what the difference is between $2m$ and $m$.  $2m$ stands for $n$, does $m$ stand for modulo?  I'd like to see and example with an even number or two, such as $4$, and $6$. 

I tried, $2m = 4$

$$0, 1, 3, 2, 2\ldots$$ 
wait, that can't be right.

It should turn out with non repeating sequences, and non recurring pairs between rows.
I can outline the other methods I've found (but don't understand) if this one is not the best.
If I understand your description correctly, for $2m=4$:
$$\begin{bmatrix}
    0&amp;  1&amp;  3&amp;  2\\
    1&amp;  2&amp;  0&amp;  3\\
    2&amp;  3&amp;  1&amp;  0\\
    3&amp;  0&amp;  2&amp;  1\\
\end{bmatrix}$$

For $2m=6$:
$$\begin{bmatrix}
    0&amp;  1&amp;  5&amp;  2&amp;  4&amp;  3\\
    1&amp;  2&amp;  0&amp;  3&amp;  5&amp;  4\\
    2&amp;  3&amp;  1&amp;  4&amp;  0&amp;  5\\
    3&amp;  4&amp;  2&amp;  5&amp;  1&amp;  0\\
    4&amp;  5&amp;  3&amp;  0&amp;  2&amp;  1\\
    5&amp;  0&amp;  4&amp;  1&amp;  3&amp;  2
\end{bmatrix}$$

edit: In each case, the first row follows the pattern you gave; in subsequent rows, each number is one more than the number above it, but wrapping around at the size of the square (e.g. in the case where $n=2m=6$, after $5$ comes $0$).

Let's look at $n=2m=12$ $(m=6)$ a bit more closely.  The pattern that you gave said:
$$0, 1, 2m - 1, 2, 2m - 2, 3,\ldots, m + 1, m$$

So, the first row is:
$$0, 1, 12 - 1, 2, 12 - 2, 3,\ldots, 6 + 1, 6$$

which is
$$0, 1, 12-1=11, 2, 12-2=10, 3, 12-3=9, 4, 12-4=8, 5, 12-5=6+1=7, 6$$

Now, the second row is:
$$0+1=1, 1+1=2, 11+1=12$$
becomes
$$0, 2+1=3, 9+1=10, 4+1=5, 8+1=9, 5+1=6, 7+1=8, 6+1=7$$

Continuing gives:
$$\begin{bmatrix}
    0&amp;  1&amp; 11&amp;  2&amp; 10&amp;  3&amp;  9&amp;  4&amp;  8&amp;  5&amp;  7&amp;  6\\
    1&amp;  2&amp;  0&amp;  3&amp; 11&amp;  4&amp; 10&amp;  5&amp;  9&amp;  6&amp;  8&amp;  7\\
    2&amp;  3&amp;  1&amp;  4&amp;  0&amp;  5&amp; 11&amp;  6&amp; 10&amp;  7&amp;  9&amp;  8\\
    3&amp;  4&amp;  2&amp;  5&amp;  1&amp;  6&amp;  0&amp;  7&amp; 11&amp;  8&amp; 10&amp;  9\\
                   &amp;&amp;&amp;&amp;&amp;\vdots
\end{bmatrix}$$
As I have essentially asked the same question somewhere else, I wanted to share the answer I was given there by Gage:


  See here for more information [on row-complete Latin squares] http://personal.maths.surrey.ac.uk/st/H.Bruin/MMath/LatinSquares.html


The reference outlines a fairly straight forward algorithm and I wrote a a quick MATLAB script to build me an order-n square: https://gist.github.com/hcc23/274b76edc3dd82d5de1a