There are two ways of doing this. One is Ross Millikan's: you will make ten "up" moves, and 20 "right" moves; the only question is which order you make them in. Imagine placing the "right" moves on a row; now you need to decide where to do the "up" moves: you do so by inserting them "in between" (or before, or after) the "right" moves. So you need to choose ten places to put "up" moves: there are 21 locations for them (nineteeen in between the "right" moves, one before all of them, one after), and you are allowed to choose the same location more than once.
This is a combinations-with-repetitions: the formula is $\binom{n+r-1}{r}$, where you have $n$ possibilities, and must make $r$ choices with repetitions allowed. In this case, $n=21$, $r=10$, so you get $\binom{30}{10}$.
There is another way of doing it, which is more graphical. I'll do it with a 4 by 3 array so you see how it works. You have this array:
$$\begin{array}{cccc}
\cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot
\end{array}$$
Now, you start at the bottom left, so there is only one way to get there; we put a $1$ next to it.
$$\begin{array}{llll}
\cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot\\
\cdot\;1& \cdot & \cdot & \cdot
\end{array}$$
Then, you can go either up or right; there is only one way to get to those points (via the first move); we put a $1$ next to them:
$$\begin{array}{llll}
\cdot & \cdot & \cdot & \cdot\\
\cdot\;1 & \cdot & \cdot & \cdot\\
\cdot\;1& \cdot\;1 & \cdot & \cdot
\end{array}$$
Now: to get to $(1,1)$, you can either get to it from $(1,0)$ or from $(0,1)$; since there is only one way to get to each of those, there are two ways to get to $(1,1)$. On the other hand, only one way to get to $(2,0)$ or to $(0,2)$:
$$\begin{array}{llll}
\cdot\;1 & \cdot & \cdot & \cdot\\
\cdot\;1 & \cdot\;2 & \cdot & \cdot\\
\cdot\;1& \cdot\;1 & \cdot\;1 & \cdot
\end{array}$$
Next: to get to $(1,2)$, you can arrive either from $(0,2)$ (one way of being there), or from $(1,1)$ (two ways of getting there); so in total, three ways. Likewise, you have three ways to get to $(2,1)$, because you can either go up from $(2,0)$, and there is only one way to do all of that, or you can go right from $(1,1)$ (and there are two ways of doing that, corresponding to the two ways there are to get to $(1,1)$; so we have:
$$\begin{array}{llll}
\cdot \;1 &\cdot\;3 & \cdot & \cdot\\
\cdot\;1 & \cdot\;2 & \cdot\;3 & \cdot\\
\cdot\;1& \cdot\;1 & \cdot\;1 & \cdot
\end{array}$$
Continuing this way, we get:
$$\begin{array}{llll}
\cdot\;1 & \cdot\;3 & \cdot\;6 & \cdot\;10\\
\cdot\;1 & \cdot\;2 & \cdot\;3 & \cdot\;4\\
\cdot\;1 & \cdot\;1 & \cdot\;1 & \cdot\;1
\end{array}$$
So there are $10$ ways to get to the top right corner in the 4 by 3 case.
You may even recognize that these numbers are just Pascal's triangle lying on its side! Well, there is a combinatorial formula for the entries of Pascal's triangle: the $r$th entry in the $m$th row corresponds to the coefficient of $a^{m-r}b^{r-1}$ in the binomial expansion of $(a+b)^{m-1}$, so it equals $\binom{m-1}{r-1}$. To figure out the entry that corresponds to the top right corner, note that you go "down" one row for each position on the $x$-axis, and another one for each step up. So here we have gone to the 4th row on the horizontal steps, and then to the 6th row on the by going up twice. Each step up is a move "right" on the row. So with a 4 by 3, we are in row $4+(3-1)=6$, and we are in position $3$ of that row. According to the formula above, that corresponds to $\binom{4+2-1}{3-1}=\binom{5}{2}$. This corresponds to the need to make $3$ moves right and two moves up, so we need choose where to place the two up moves among the three right moves; there are four places to put them in (before the three, after the three, or in the two spaces in between), so the formula I gave above gives this answer as well.
Well, it's easy to say something about the case $c=2$. There is no strategy, since all you can do is alternate between the two colors. Since there is a path of length $2n-2$ (counting the edges) between any two points, no game can take more than $2n-2$ moves. It's easy to construct a game that takes $2n-2$ moves; just use a checkerboard coloring.
At the other extreme, it's easy to say something about the case $c=n^2$. If each node is a different color, then you can always extend by one node at each move, and you can never extend by more than one node, so an optimal strategy will take $n^2-1$ moves.
So now we just have to fill in all the cases in between....
Best Answer
This seems like a job for dynamic programming. This means that we do the problem backwards in a sense. In any case where both players can move in the same direction, we obviously want to move in that direction. When we have a choice of two directions to move, we want to move in the direction that leaves the position requiring fewer steps.
Therefore, we start by computing the minimum number of steps required to move from positions where the players are close together, and keep track of the results. Then we increase the distance by $1$, and use the saved values to compute the new move requirements.
We can modify this to keep track of what the best move is in each case.
I wrote the following python script to do this.
In the green path, X and Y are just placeholders for the starting positions, and U,D,L,R are the moves needed to get to that cell. The red path is the same as the green, with the directions reversed.
The program produces the following output:
So, if I haven't made any mistakes, $26$ steps are required, the first $4$ of which are "up", and so on.
I haven't checked this script carefully at all. You should probably make some small cases that you can solve by hand, and test the script on them. Let me know of any problems, please.