In statistics the third moment is used to calculate skewness. I would guess this has a physical analogy. Although I haven't thought this through, I'd guess it would be possible to take a disk and deform it asymmetrically so that the centre of mass and moment of inertia remained the same, but the third moment changed. In this case the third moment would be telling you about the asymmetry of the disk.
This is a direct consequence of what comes out when you go from equations of motion for single point particles to collective motion of extensive bodies. You start with the second Newton's law:
$$m_i \vec{a}_i = \vec{F}_{i}$$
Summation over the entire system cancels out pairs of mutually opposite internal forces, leaving us with external force and acceleration:
$$\sum m_i \vec{a}_i = \vec{F}$$
Now, by definition, it's $\vec{a}_i = \frac{d^2}{dt^2}\vec{r}_i$. So this can be written as
$$\frac{d^2}{dt^2} \sum m_i \vec{r}_i = \vec{F}$$
notice the sum is exactly the thing that you replace with $m\vec{r}_0$ where $m=\sum m_i$ is the total mass and $\vec{r}_0$ is the center of mass. Why we use the center of mass? Because it turns out equations for linear motion reduce to FORCE = TOTAL MASS * ACCELERATION OF THE CENTER OF MASS.
Now for the rotational part. Of course, if an equation is true, it remains true if you do the same operation on both sides. In this case, according to our (arbitrarily chosen) coordinate frame, the positions of particles were $\vec{r}_i$. Just cross Newton's law with this:
$$m_i \vec{r}_i\times \vec{a}_i = \vec{r}_i\times \vec{F}_{i}$$
Now perform summation over the entire body ... again. In this case, the right-hand side separates into torque of external forces, plus the summation over all pair interactions:
$$\sum_i m_i \vec{r}_i\times \vec{a}_i = \sum_i \vec{r}_{ext,i}\times \vec{F}_{ext,i}+\sum_{i,j}(\vec{r}_i-\vec{r}_j)\times \vec{F}_{ij}$$
Here I took into account that pairs of internal forces are equal in magnitude, but as the positions are different, they wouldn't necessarily cancel out. But classically, they do: forces of one particle on another point along their connecting line so the cross products of the second term are zero (magnetism requires caution here, but I won't get into it).
What's left is
$$\sum_i m_i \vec{r}_i\times \vec{a}_i = \vec{M}$$
You have $\vec{a}_i=d^2 \vec{r}_i/dt^2$. Also, observe:
$$\frac{d}{dt}(\vec{r}\times\frac{d}{dt}\vec{r})=\frac{d}{dt}\vec{r}\times\frac{d}{dt}\vec{r}+\vec{r}\times\frac{d^2}{dt^2}\vec{r}$$
The first term is zero (cross a vector with itself), so you can write
$$\frac{d}{dt}\sum_i m_i \vec{r}_i\times \vec{v}_i = \vec{M}$$
This is the conservation of angular momentum. However... if you have a rigid motion, then you can write the velocities as a sum of translations and rotations! Let's say that we have some arbitrary reference point $\vec{A}$ and a displacement from it: $\vec{r}_i=\vec{A}+\vec{R}_i$. Now, the time derivative must take into account motion of the reference, but also consider rotation about the reference:
$\vec{v}_i=\vec{v}_A+\omega\times\vec{R}_i$. Here the second term is the tangential velocity of rotation around the axis $\omega$. Now, we put this back into the above equation and express all with the relative displacement:
$$\frac{d}{dt}\sum_i m_i (\vec{A}+\vec{R}_i)\times (\vec{v}_A+\omega\times\vec{R}_i) = \vec{M}$$
$$\frac{d}{dt}\left(m\vec{A}\times \vec{v}_A+\vec{A}\times\omega\times\sum_i m_i \vec{R}_i+\sum_i m_i \vec{R}_i\times \vec{v}_A+\sum_i m_i \vec{R}_i\times \omega\times\vec{R}_i\right) = \vec{M}$$
This looks horrible, and it is, unless you specifically put the reference into the center of mass, $\vec{A}=\vec{r}_0$. In that case, the relative positions are expressed in system where $\sum_i m_i \vec{R}_i=0$. The thing reduces further, and let's also recall the double cross product expansion:
$$\frac{d}{dt}\left(m\vec{r}_0\times \vec{v}_0+\sum_i m_i\left( \omega|\vec{R}_i|^2-\vec{R}_i(\omega\cdot\vec{R}_i)\right)\right) = \vec{M}$$
The first term is the angular momentum of the center of mass motion. The second term is $J\vec{\omega}$, where $J$ is the tensor defined as
$$J_{ab}=\sum_{i} \left(|R_i|^2\delta_{ab}-R_{i,a}R_{i,b}\right)$$
In short... the second Newton's law and subsequently conservation of momentum reduces in the case of rigid body motion to an expression that involves the second moment of mass.
Best Answer
Given some distribution or density $\rho(x),$ a moment is the 'expectation value' of some power of $x \in \mathbb{R}$. To be precise, the $n$-th moment $M_n$ is given by $$M_n = \int_{\mathbb{R}} x^n \rho(x) \mathrm{d}x.$$ In the mechanics case, $\rho(x)$ is simply the mass density.
You can extend this to vectors in $\mathbb{R}^d$ in a straightforward way; for example, for the moment of inertia you replace $x^2$ by $\mathbf{x}^2 = x_1^2 + \ldots x_d^2$ to obtain $$I = M_2 = \int_{\mathbb{R}^d} \mathbf{x}^2 \rho(\mathbf{x}) \mathrm{d}^dx$$ which should match the definition given in your mechanics textbook.
For the first moment of mass, you need to distinguish different directions. As you indicate, you can choose your coordinates such that
$$\int_{\mathbb{R}^d} x_i \,\rho(\mathbf{x}) \mathrm{d}^d x = 0$$ where $i$ runs over the coordinates. In three dimensions, you have $x_1 = x, x_2=y$ and $x_3=z.$