Subtraction of inequalities

First thing you can do is invoke the triangle inequality, i.e. $|x + y| \leq |x| + |y|$. Note that this equality works for negative $x$, $y$ as well. This gives you $$ \begin{eqnarray} |k_1 - k_2| &\leq& |k_1| + |k_2| &\leq |a| + |b| \text{ and since }k_1-k_2 \leq |k_1-k_2| \\ k_1 - k_2 &\leq& |k_1| + |k_2| &\leq |a| + |b| \end{eqnarray} $$

For arbitrary $k_1 \leq a$, $k_2 \leq b$, you can't do better than that, but since you stated that $k_1,k_2 \geq 0$, you can. $k_1,k_2$ being positive means that $$ \begin{eqnarray} -b &\leq& -k_2 &\leq& k_1 - k_2 &\leq& k_1 &\leq& a\\ -a &\leq& -k_1 &\leq& k_2 - k_1 &\leq& k_2 &\leq& b \end{eqnarray} $$ because subtracting a positive number can only make a number smaller, never bigger.

Let's look at some example. We know that $$ 4<8, 2<10, $$ but we don't have $4-2<8-10$. On the other hand, $$ 3<9,4<8$$ but you cannot say $3-4>9-8$. There is nothing to deduce from substrating inequalities that are of the same side