If the mean and median underestimate the true central tendency, why use them? In skewed distributions, both the mean and the median can easily underestimate or overestimate the true central tendency. For example, have a look at this violin plot:

The median is shown here in red. It seems to overestimate the true central tendency. This is the same for the mean. The best central tendency description would be the mode of that distribution (which would get the peak).
This is a basic question: but why use mean and median for skewed distributions? Do people even use them? Wouldn't it make sense to get the maximum point of the Kernel density estimate?
 A: If there is a mistake in your assumptions, it is that these are the only measures of central tendency available, e.g., the "mean" refers only to the arithmetic mean. In fact, there are many, many measures of central tendency such as the Pythagorean trio -- arithmetic, geometric and harmonic means -- Hodges-Lehmann estimators as well as, yes, kernel density estimates, to name just a few. Not to mention that each and every distribution, when the moments exist, has a formula defined and assigned for calculating a "mean" appropriate for that distribution. That said, there are also distributions such as Tweedie's, for which the "mean" as such is infinite and does not exist.
A: Expected values of calculations
Distribution mean is useful to calculate expected total results - for example, if you want to estimate the total weight of 100 plane passengers, or the total expected return of a portfolio of many investments, then the mean of that distribution will be useful to you and the mode will be rather useless.
In cases where you don't really care about the actual individual values you get, but about their sum, product or other aggregate, then you need a measure that is consistent within that aggregate function - arithmetic mean for sums, geometric mean for products (e.g. for growth rates), etc.
A: 
In skewed distributions, both the mean and the median can easily
  underestimate or overestimate the true central tendency. [...] Wouldn't
  it make sense to get the maximum point of the Kernel density estimate?

No it wouldn't, at least not always. 
Take as an example the exponential distribution: it is parametrized by $\lambda$, its expected value is $\lambda^{-1}$, the same as its mean, its median is $\lambda^{-1} \ln(2)$ and its mode is always $0$, regardless of parametrization. So all the values from this distribution are greater than or equal to the mode -- what does that tell us? Not much... 
It is the opposite with the mean which marks the probability mass center; the median also provides us with similar information. Yes, $0$ is the most likely value, but we are more interested here in the tail of the distribution (since it's a tail-only distribution). 
Finally, the mode would be a useless summary statistic if you would like to compare different exponential variables.
Check out also: If mean is so sensitive, why use it in the first place?
