Why is kurtosis compared to a normal distribution? Why is the kurtosis equation compared to a normal distribution? Are there cases where you would want to compare the tails versus some other types of distributions? 
$$\gamma = \frac{M_4}{\sigma^4} -3$$
In the above equation would you simply replace the $3$ with the kurtosis of another distribution with which you would like to compare?
 A: Subtracting $3$ has at least one other justification besides taking the normal distribution as the standard.  The functional $\kappa_4$ given by
$$
\kappa_4(X) = \operatorname{E}((X-\mu)^4) - 3(\operatorname{var}(X))^2, \quad \text{where } \mu = \operatorname{E}(X)
$$
is the fourth cumulant of the distribution of $X$.  It is


*

*translation-invariant, i.e. $\kappa_4(X+c) = \kappa_4(X)$ if $c$ is constant (i.e. not random);

*homogeneous of degree $4$, i.e. $\kappa_4(cX) = c^4\kappa_4(X)$;

*"cumulative", i.e. $\kappa_4(X_1+\cdots+X_n) = \kappa_4(X_1)+\cdots+\kappa_4(X_n)$ if $X_1,\ldots,X_n$ are independent.


That last property holds only if the coefficient $-3$, rather than some other number, appears where it does.
A: I think there are two conventions out there on what to call your $\gamma$: the first is "kurtosis" and the second is "excess kurtosis". The second term is more precise as it makes it clear that the Gaussian distribution is the benchmark. Also, the second convention implies that the "raw" kurtosis (without the -3 term) would not refer to any specific benchmark. So one answer is yes, you could make up your own "excess kurtosis with respect to distribution XYZ" by replacing the 3 with the kurtosis of that distribution. But in order to avoid confusion it's probably better to keep the Gaussian benchmark intact.
A: Sure, you could use any distribution for comparison. Higher kurtosis for your distribution vs. the comparator implies your distribution has greater tail weight (or tail leverage; see below) than the comparison distribution. Lower kurtosis for your distribution vs. the comparator implies your distribution has less tail weight (leverage). 
Logic is as follows. Consider your random variable $X$, having distribution $p_X(x)$. Assume finite fourth moment and let $V = \{(X-\mu)/\sigma\}^4$. Then the (non-excess) kurtosis of $X$ is $\kappa =  E(V)$. 
A standard way of understanding expectation is the "point of balance" of the distribution. Thus the distribution $p_V(v)$ balances at $\kappa$, which is the kurtosis of your distribution $p_X(x)$. 
Now, suppose your comparator distribution has kurtosis $\kappa_0$. Locate $\kappa_0$ on the horizontal axis of the graph of  $p_V(v)$ with a fulcrum. If  $p_V(v)$ falls to the right of the fulcrum at $\kappa_0$, then $\kappa > \kappa_0$ and your distribution $p_X(x)$ is heavier-tailed than the comparator.  If $p_V(v)$ falls to the left of the fulcrum at $\kappa_0$, then $\kappa < \kappa_0$ and your distribution $p_X(x)$ is lighter-tailed than the comparator.   
Greater "heaviness" of tail refers to greater leverage of the tail rather than greater mass in the tail.  You can have less mass in the tail with greater leverage (higher kurtosis), provided the mass is sufficiently distant from the mean. 
