Upper bound on the average number of overlaps for an interval within a set of intervals
If we ignore $L$ and focus only on the parameters $C,K,M$, the following upper bound is asymptotically tight, i.e., it's about the best you can do, up to a constant factor:

$$\text{Avg}(\mathcal{I}) \le \min(MC,K).$$



Proof that it's an upper bound: Fix any interval.  We're promised that it overlaps with at most $K$ others.  Also, the interval has at most $C$ points in it, so by a union bound over those points, we can also infer it overlaps with at most $MC$ others.  Therefore, it overlaps with at most $\min(MC,K)$ others.  Now the average of a bunch of numbers that are all $\le \min(MC,K)$ will itself be $\le \min(MC,K)$.



Proof that it's tight: I will show the construction of a set of intervals where $\text{Avg}(\mathcal{I}) \sim \min(MC,K)/4$.  There are two cases:


Case 1: $K \ge MC$.  Then use $M/2$ copies of each interval of the form $[i,i]$ (i.e., each length-0 interval), and use $M/2$ copies of the interval $[1,C]$.  You can observe that the latter each intersect with $\sim MC/2$ other intervals (and all intersect with fewer than $K$).  So, the average is about $MC/4$.  This gives a collection of intervals where each point intersects with $M$ intervals, each interval intersects with $\le K$ others, and $\text{Avg}(\mathcal{I}) \sim MC/4 = \min(MC,K)/4$.
Case 2: $K &lt; MC$.  Set $C' = K/M$, apply the above construction to the interval $[1,C']$ instead of $[1,C]$, and we obtain a collection of intervals where each point intersects with $M$ intervals, each interval intersects with $\le K$ others, and $\text{Avg}(\mathcal{I}) \sim MC'/4 = K/4 = \min(MC,K)/4$.




If you also care about the dependence on $L$,  you might be able to build on the above analysis to see how it might depend on $L$.