First, some quick notes (based on the earlier answers and the comments):
Given a vector $\vec v$,
we can compute its
square-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 $$\tilde P\cdot \tilde P=m^2c^2\quad\fbox{$\frac{1}{c^2}$Square-Energy of Rest Mass $m$}\mbox{ or } \fbox{$\frac{1}{c^2}$Square-Energy of [Invariant] Mass} $$
