Here's a diagram of neutron decay.
Up and down quarks have rest masses of 2-4 MeV. The $W$ boson has a rest mass of 80 GeV.
Where has this extra mass come from?
[Physics] Where does $W$ boson mass come from in neutron decay
conservation-lawsfeynman-diagramsmassstandard-modelvirtual-particles
Related Solutions
The extra mass of the proton and neutron is not due to chiral symmetry breaking. It is due to the energy in the electromagnetic and strong force fields.
If chiral symmetry were an exact symmetry of the Lagrangian then the pions and other mesons (not the baryons) would have zero mass due to spontaneous chiral symmetry breaking. The chiral symmetry is not exact due to the small bare masses of the quarks so the mesons are not exactly massless.
From this you can see that your question is a bit confused, but one part that is correct is that most of the mass in ordinary atomic matter is not due to the Higgs mechanism. When people say that the Higgs boson will give us information abut the origin of mass they mean the bare masses of non-composite particles such as electrons and quarks.
You say:
Now, when we talk about energetically favourably bound systems, they have a total mass-energy less than the sum of the mass-energies of the constituent entities.
and this is perfectly true. For example if we consider a hydrogen atom then its mass is 13.6ev less than the mass of a proton and electron separated to infinity - 13.6eV is the binding energy. It is generally true that if we take a bound system and separate its constituents then the total mass will increase. This applies to atoms, nuclei and even gravitationally bound systems. It applies to quarks in a baryon as well, but with a wrinkle.
For atoms, nuclei and gravitationally bound systems the potential goes to zero as the constituents are separated so the behaviour at infinity is well defined. If the constitiuents of these systems are separated to be at rest an infinite distance apart then the total mass is just the sum of the individual rest masses. So the bound state must have a mass less then the sum of the individual rest masses.
As Hritik explains in his answer, for the quarks bound into a baryon by the strong force the potential does not go to zero at infinity - in fact it goes to infinity at infinity. If we could (we can't!) separate the quarks in a proton to infinity the resulting system would have an infinite mass.
So the bound state does have a total mass less than the separated state. It's just that the mass of the separated state does not have a mass equal to the masses of the individual particles.
You can look at this a different way. To separate the electron and proton in a hydrogen atom we need to add energy to the system so if the added energy is $E$ the mass goes up by $E/c^2$. As the separation goes to infinity the energy $E$ goes to 13.6eV. If we try to separate the quarks in a proton by a small distance we have to put energy in and the mass also goes up by $E/c^2$ just as in any bound system. But with the strong force the energy keeps going up as we increase the separation and doesn't tend to any finite limit.
Best Answer
You have drawn a Feynman diagram.
Feynman diagrams are iconic shorthand for integrals over the variables of the problem. The calculation gives the probability for the reaction to happen, in this case the decay of a neutron .
The observables are the four vectors of the initial (neutron) and final particles. The integral is over the variables .
Here is a simpler labeled diagram
This is the expression that has to be integrated over the limits of the variables.
The incoming lines represent real particles, and also the outgoing lines. The line in between represents the functions under the integral. This line has to carry the charge and quantum numbers that conservation laws impose. In addition there exists a function under the integral, called a propagator, which has in the denominator the mass of the named particle. In the case above it is the photon's zero mass.
Under the integral the four vector of this "photon" line cannot have zero mass because of the spread of the variables of integration, so it is off mass shell and called a virtual photon.
In your diagram for neutron decay the corresponding denominator is ((p-q)^2-m_W^2). The large mass is crucial and represents together with the coupling constant, the "weakness" of the interaction. That is why the internal line is identified with the W. It has all the quantum numbers but a variable mass of the four vector it represents. It is called virtual for this reason.
The W boson mass comes within the integral represented by the diagram, in the denominator of the propagator. The line represents an off mass shell virtual W.