When proving that $q\text{-}\textsf{COL} \leq_m^P q\text{-}\textsf{COL}_{2q-1}$, what you are really showing is that checking whether a graph with maximal degree $2q-1$ is $q$-colorable isn't any easier than checking whether an arbitrary graph is $q$-colorable. Why is this an interesting statement? Since if we replace $2q-1$ with a small enough number, then checking whether a graph is $q$-colorable does become easier. For example, any graph with maximal degree $q-1$ is $q$-colorable, so checking whether such a graph is $q$-colorable is easy, whereas checking whether an arbitrary graph is $q$-colorable is NP-hard for all $q \geq 3$.
How do you show that $q\text{-}\textsf{COL} \leq_m^P q\text{-}\textsf{COL}_{2q-1}$? You need to transform a given graph $G$ to a graph $H$ with maximum degree $2q-1$ such that $H$ is $q$-colorable iff $G$ is $q$-colorable. The most natural way of doing this is by splitting high-degree vertices. I'll illustrate how to do it for the case $q=2$ (which in reality isn't so interesting — why?), and will let you generalize the construction. Once you do that, you'll also see where $2q-1$ is coming from.
Suppose therefore that $q = 2$. First of all, we number the neighbors of all vertices. For each edge $(v,w)$, there are numbers $i,j$ such that $w$ is the $i$th neighbor of $v$, and $v$ is the $j$th neighbor of $w$. Now, we replace each vertex $v$ of degree $d$ with a path $v_0,\ldots,v_{2d}$, and connect $v_{2i}$ and $w_{2j}$ if $w$ is the $i$th neighbor of $v$ and $v$ is the $j$th neighbor of $w$. Note that all vertices now have degree at most 3. The construction works basically since the vertices $v_0,v_2,\ldots,v_{2d}$ all get the same color — I'll let you work out the details.
