Difference between: 'Rest Mass' length, 'Centre of Mass' length and 'Invariant Mass' length in special relativity and particle physics? Firstly, after reading this source, the length, $s$ of an energy-momentum four-vector is 
$$s=m_0c^2\quad\fbox{Rest Mass}$$
But, according to this, the length of an energy-momentum four-vector is $$s=m_T^2c^4=E_T^2-p_T^2c^2\quad\fbox{Centre of Mass}$$
From yet another source, the length of an energy-momentum four-vector is $$s=m^2c^2\quad\fbox{Invariant Mass}$$ 

Basically, when I take the 'length' (or norm) of a four-vector, I'm not sure which one to use. All of them are invariant according to their sources. My confusion is mainly between the last two lengths.
So put simply, which length do we use and when? What is the difference between them?

Remark: Before I wrote this, I looked at answers to this question, but, I don't think it answers my question here.
 A: 
[T]he length of an energy-momentum
  four-vector is $s=m_0c^2$ $\quad\fbox{Rest Mass}$

Ok, this is true but highly condemnable language. The concept of rest mass has been banished from physics alongside the concept of relativistic mass for very good reasons. There is only mass which is given by the length of the four-momentum vector (up to a factor of $c^2$). In old school notation, there was a relativistic mass $m_v$ which was defined as $E/c^2$ and thus, there was also rest mass $m_0$ which was related to the energy of the particle in its rest frame. In other words, $m_0=E_0/c^2$. As you can notice, in the rest frame, the four-momentum vector takes the form $(E_0,0,0,0)$ and thus, $m_0c^2$ would be the length of the vector (this would be the length in all frames even if we calculated it in rest-frame because lengths are invariant). So, that's why the reference says that rest mass is the length of the four-momentum vector. But, in modern language, we call the same quantity just mass. And since it is also invariant, we also sometimes call it invariant mass. 

[T]he length of an energy-momentum four-vector is
  $s=m_T^2c^4=E_T^2-p_T^2c^2\quad\fbox{Centre of Mass}$

Yes, this is true. This refers to a system of particles. When you have a system of particles, the analog of going to the rest frame of a particle is going to the frame where the total $3-$momentum is zero. As you can see in the source you linked, this is what is defined as the center of mass frame. So, the total energy in the center of mass frame would give you the mass of the composite system, or, in other words, the length of the four-momentum vector of the composite system.

[T]he length of an energy-momentum four-vector is
  $s=m^2c^2\quad\fbox{Invariant Mass}$

Assuming, it is supposed to mean $m^2c^4$ and not $m^2c^2$, this is absolutely correct and the correct version of the first formula. Since we don't use the rest mass / relativistic mass terminology anymore, there is just mass. And the quantity that was called $m_0$ back in the day is now just called $m$, the mass. So, this is actually the same formula as the first one but in better notation. 
So, in summary, all three formulas are correct. The first is in bad notation, the last is perfect, and the middle one is designed to explicitly talk about a composite system. 

Edit
I notice there is some issue with the factors of $c$ in the formulas you wrote. I am ignoring them as they are trivial, in the sense that you can figure out the correct factors of $c$ by dimensional analysis. Also, some formulas are talking about the squared length and the others about length. There is nothing deep there, you can talk about either, the squared length would correspond to mass-squared (times $c^4$) and the length would correspond to mass (times $c^2$). However, it is important to notice that in what you cite, the same symbol $s$ is being used for both length and squared-length which can be confusing. To be explicitly clear, the first formula $s=m_0c^2$ gives you the length, the second formula $s=E_T^2-p_T^2c^2$ gives you squared length, and the third formula $s=m^2c^4$ gives you squared length.  The widespread convention is to use $s$ for squared length, for example, see Mandelstam variables.
A: First, some quick notes (based on the earlier answers and the comments):

*

*Given a vector $\vec v$,
we can compute itssquare-magnitude or square-norm as
$\vec v\cdot \vec v$.
Its magnitude (or norm or length) is $\left|\vec v\right|=\sqrt{\vec v \cdot\vec v}.$


*In the OP, there are inconsistencies in conventions [using factors of $c$ (see the next bullet point)] and inconsistencies in referring to each item as a length (or magnitude) with the symbol $s$. To fix the confusions, I will use the symbol $\ell$ for "length" [in units of energy].

Using the conventions of the first item,
$$\ell=m_0c^2\quad\fbox{ Energy-equivalent of Rest Mass},$$
the second item's source was misread [it says
"$s$ is the square of the centre-of-mass energy"]
so that this second item
should be revised to read
$$\ell_T=\sqrt{s}=\sqrt{m_T^2c^4}=m_Tc^2=\sqrt{E_T^2-p_T^2c^2}\quad\fbox{Energy-equivalent of the Centre of Mass},$$
and
the third item's source was misread and also uses the "length" in units of momentum
so that third item
should be revised to read
$$\frac{1}{c}\ell=mc\quad\fbox{Momentum-equivalent of Invariant Mass}$$


*There are conventions involving the speed of light $c$.
Special relativity uncovered previously unrecognized relationships among mass, momentum, and energy, which was historically expressed with different units. The speed of light plays a role as a unit-conversion factor. (This is one of the lessons from "The Parable of the Surveyors" from Spacetime Physics by Taylor & Wheeler, where, for some reason, North-South distances (measured in miles) and East-West distance (measured in meters) were treated differently until they were seen as components of a vector, which is best described with common units.)

So, the 4-momentum can be expressed in various unit-conventions:
momentum-convention: $\tilde P=(E/c,p_x,p_y,p_z)$ with "length" $(mc)=\sqrt{(E/c)^2-p^2}$
energy-convention: $\tilde P=(E,p_xc,p_yc,p_zc)$ with "length" $(mc^2)=\sqrt{E^2-(pc)^2}$
mass-convention: $\tilde P=(E/c^2,p_x/c,p_y/c,p_z/c)$ with "length" $m=\sqrt{(E/c^2)^2-(p/c)^2}$

To answer the bountied question "which length to use and when?"
more fully, I offer this update to my original answer, which is preserved below this update.
In the OP's order [with terms revised here],
the first and third item are essentially the same,
but "rest" only applies to a particle with positive invariant-mass.
The second item applies to a "collection of particles" (which may be a single particle or several particles).

Consider the decay of a particle C into two identical-mass particles A and B. (In this section, all particles are treated as point-particles
and this interaction occurs at one event in spacetime.)
Each particle has a 4-momentum vector.
I will denote the 4-momentum of a particle A by $\tilde A$, etc.
In addition,
it is useful to define
the 4-momentum of a collection of particles [meeting at an event]

as the sum of the individual particle 4-momenta,
as if it were a single fictitious particle
with that 4-momentum.
(This is akin to locating the center of mass of an object at a
location where there may be no mass located, like the center of a hoop.)
As a special case, one could define the 4-momentum of our system
before and after the decay or collision:
$$\tilde P_{before}=\tilde C,$$
and
$$\tilde P_{after}=\tilde A + \tilde B$$
which can be suitably generalized for more particles.
(Furthermore, if desired, one could define, for example,
a 4-momentum for a subset of particles after the collision.)
By conservation of total-momenta in this system,
\begin{align*}
\tilde P_{before} &= \tilde P_{after}\\
\tilde C &=\tilde A +\tilde B,
\end{align*}
which could be described by drawing a polygon in
an energy-momentum diagram. The diagram is meant to
display and facilitate calculations involving 4-momenta.
For concreteness, let particle C have [invariant rest-]mass $20$,
and let particles A and B, each have [invariant rest-]mass $6$.
Henceforth, I will drop the prefix "[invariant rest-]".
In C's frame, A has velocity $8/10$ and B has velocity $-8/10$.

I draw my diagram on "rotated graph paper"
so that we can more easily count "tickmarks" of mass.

(ref: Relativity on Rotated Graph Paper,
American Journal of Physics 84, 344 (2016);
https://doi.org/10.1119/1.4943251)
Using the metric, we can write the square-norm of $\tilde C$ (or square-magnitude of $\tilde C$)
by
$$\tilde C\cdot \tilde C,$$
an invariant which can be
interpreted [up to conventional factors $c$ and signature-conventional signs]
as the square of the invariant-mass of the particle $C$.
(Typically, we expect that these square-norms are [in my conventions]
non-negative.)
I will adopt conventions so that
$$m_C^2 \equiv \tilde C \cdot \tilde C.$$

So, in my example, the squares of the invariant-masses of the particles are
$$m_C^2=(20)^2=400 \qquad m_A^2=(6)^2=36 \qquad  m_B^2=(6)^2=36.$$
These can be interpreted as the areas of their respective "mass diamonds"
where the 4-momentum vector is a timelike diagonal of the diamond (as shown for $\tilde A$)

*

*If a square-norm is positive [in my convention], then particle is said to be massive
and thus, one can refer to the magnitude of the 4-momentum vector

$$m_C=\left|\tilde C\right| =\sqrt{\tilde C \cdot \tilde C}$$ as the so-called $\fbox{rest-mass of particle C}$.
Since $\tilde C \cdot \tilde C$ is invariant, then $\sqrt{\tilde C \cdot \tilde C}$
is also an invariant, and one might hear $m_C$ referred to as the [invariant rest-]mass.

So, in my example, the invariant rest-masses of the particles are
$m_C=(20)$, $m_A=(6)$, and $m_B=(6)$. And thus, the 4-momenta can expressed in terms of their respective invariant-rest-masses and their [unit-timelike] 4-velocities:
$$\tilde C = m_C \frac{\tilde C}{m_C}= m_C \hat C \qquad  \tilde A = m_A \hat A \qquad \tilde B = m_B \hat B $$



*If a square-norm of a nonzero 4-momentum vector is zero,
then particle is said to be massless, and the vector is lightlike.
Suppose this 4-momentum vector is $\tilde K$
(as might appear in the decay of a neutral pion into two photons:
$\tilde \pi_0 = \tilde K_1 + \tilde K_2$).
Since there is no rest-frame for a lightlike particle with 4-momentum $\tilde K$ (i.e. no unit-timelike vector parallel to $\tilde K$),
one may wish to just refer to $$m_K=\sqrt{\tilde K \cdot \tilde K}$$
as the $\fbox{invariant mass of particle K}$, omitting "rest" because it is not appropriate.

Thus, $m_K=(0)$.

We could also refer to $m_C$ as the "invariant mass of particle C" (and omit "rest" so that we could use "invariant mass" for both the timelike and lightlike cases.)

$\color{red}{NOTE}$: In the OP, the quantity next to the "invariant mass" box is
actually the "square of the invariant mass [times $c^2$]".
So, the box should really say the "square of the invariant mass".


These notions can be extended to "collections of particles",
for example, the "particles after the collision" in my example.
$\tilde P_{after}=\tilde A + \tilde B$.
Thus, the "square-of invariant-rest-mass of the possibly-fictitious-particle"
with 4-momentum $\tilde P_{after}$ is
\begin{align*}
\tilde P_{after}^2 &=\tilde P_{after}\cdot \tilde P_{after}\\
 m_{(A+B)}^2 &=(\tilde A+\tilde B)\cdot (\tilde A+\tilde B),\\
\end{align*}
thus,
$$m_{(A+B)} =\left| \tilde P_{after} \right|
=\sqrt{(\tilde A+\tilde B)\cdot (\tilde A+\tilde B)}$$
is the "invariant-rest-mass of the possibly-fictitious-particle".
Note that there is no particle $C$ after the decay.
Since
\begin{align*}
\tilde P_{after}
&=\tilde A + \tilde B \\
&= m_{(A+B)}\frac{(\tilde A + \tilde B)}{m_{(A+B)}}\\
&= m_{(A+B)}\frac{(m_A\hat  A + m_B\hat B)}{m_{(A+B)}}\\
&= m_{(A+B)}\hat P_{(A+B)} \\
\end{align*}
where
$\hat P_{(A+B)}$ is the 4-velocity of the "center of momentum frame"
(akin to "center of mass frame"),
then
$m_{(A+B)}$ is sometimes referred to as the
$\fbox{invariant-mass of the center of momentum frame [after the decay]}$.

$\color{red}{NOTE}$: In the OP, the quantity next to the "center of mass" box is
actually the "square of the invariant-mass of the center-of-mass frame[times $c^2$]".
So, the box should really say the "square of the invariant mass".

[original answer]
I am going to use alternative symbols and terms for clarification:

*

*the length or magnitude, $|\tilde P |$, of an energy-momentum four-vector $\tilde P$ is
$$|\tilde P|=\sqrt{ \tilde P \cdot \tilde P } =m_0c^2\quad\fbox{Energy of Rest Mass $m_0$ }\mbox{ or } \fbox{Energy of [Invariant] Mass} $$


*The square-magnitude $(\tilde P_T \cdot \tilde P_T)$ of the sum of energy-momentum four-vectors $\tilde P_T= \tilde P_1+\tilde P_2 +\ldots +\tilde P_n $ is
$$s=(\tilde P_T \cdot \tilde P_T)=m_T^2c^4=E_T^2-p_T^2c^2\quad\fbox{Square-Energy of the Centre of
   Mass [frame]}$$
The 4-momentum $\tilde P_T$ is that of the "center of mass (center of momentum frame)".
Its magnitude $|\tilde P_T|$ is called the "invariant mass of the system of particles".
Here the symbol "$s$" is a "Mandelstam variable"(https://en.wikipedia.org/wiki/Mandelstam_variables) for the square-magnitude of $\tilde P_T$ (a quadratic quantity without the ${}^2$-exponent... [not to be confused with the "$s$" in $s^2$ or $ds^2$ when discussing the square-interval between two events].



*the square-magnitude $\tilde P\cdot \tilde P$  of an energy-momentum four-vector $\tilde P$ is $$\tilde P\cdot \tilde P=m^2c^2\quad\fbox{$\frac{1}{c^2}$Square-Energy of Rest Mass $m$}\mbox{ or } \fbox{$\frac{1}{c^2}$Square-Energy of [Invariant] Mass} $$
