If I understand the question right, then $k$ is simply the number of divisors $d$ of $\text{lcm}(a,b)$ such that $a\le d\le b$ and $d\mid\gcd(a,b)$. (So in finding a longest chain, we may assume that $a$ and $b$ are relatively prime.)

In your example, a longest chain would be \begin{equation} 7200\to 7560\to 7776\to 9072\to 9720\to 10080\to 10800\to 12600\to 12960\to 13608 \end{equation} I doubt that the length can be determined without some sort of actually computing the divisors. One can interpret these divisors (via their exponents) as lattice points in a polytope, so if $a$ and $b$ have lots of divisors, one probably better counts these lattice points.

If I understand the question right, then $k$ is simply the number of divisors $d$ of $\text{lcm}(a,b)$ such that $a\le d\le b$ and $d\mid\gcd(a,b)$. (So in finding a longest chain, we may assume that $a$ and $b$ are relatively prime.)

In your example, a longest chain would be \begin{equation} 7200\to 7560\to 7776\to 9072\to 9720\to 10080\to 10800\to 12600\to 12960\to 13608 \end{equation} I doubt that the length can be determined without some sort of actually computing the divisors. One can interpret these divisors (via their exponents) as lattice points in a polytope: Suppose that $a$ and $b$ are relatively prime, and $p_1,p_2,\dots,p_r$ are the primes dividing $ab$. Then $k$ is the number of integral $e_i$ such that \begin{equation} \log a\le\sum_{i=1}^re_i\log p_i\le\log b. \end{equation} Taking the volume of the polytope as an approximation of the number of the lattice points in this polytope, we see that $k$ is about (in a vague sense of course) \begin{equation} \frac{1}{r!\prod\log p_i}((\log b)^r-(\log a)^r). \end{equation}