The mass always means the same thing – but in different theories, one uses different equations and other tools to express the mass.
Inertial mass $m$ is the quantity expressing "resistance of the object with respect to acceleration", i.e. the coefficient that enters Newton's $F=ma$. The gravitational mass is what enters the formula for the gravitational force $F=GmM/r^2$. These two masses are the same, up to a universal ratio that may be chosen to be one in good units. This equality of the two masses is known as the "principle of equivalence".
In Newtonian physics, the mass of individual particles is basically conserved. Surely the total one is conserved. In special relativity, only the "total relativistic mass" which includes the enhancement from the kinetic energy is conserved. This conservation law becomes the same law with the energy conservation law; the mass may be "converted" to energy and vice versa. The total mass and the energy are related by $E=mc^2$ and this "unified" quantity is conserved whenever the laws of physics are time-translationally invariant (Noether's theorem).
For a given particle, in special relativity, we usually use the term "mass" for the rest mass $m_0$, the mass measured in the rest frame. It's related to the energy and momentum by
$$ E^2 = m_0^2 c^4 + p^2 c^2 $$
The actual working theory used to describe practical particle physics is called quantum field theory (QFT). It's a combination of classical field theory (like Maxwell's electromagnetic fields) and the postulates of quantum mechanics.
In QFT, particles are created and annihilated by fields, they are quanta of these fields. Each particle has a value of the mass and it's encoded in the "mass term". For example, for a scalar boson, the Lagrangian is
$$ L = \frac{1}{2} \partial^\mu \Phi \partial_\mu \Phi - \frac {m_0^2}{2} \Phi^2 $$
Why is it the same mass as I defined at the beginning? It's because the field equations resulting from the Lagrangian above are the Klein-Gordon equation
$$(-\square - m_0^2 )\Phi =0$$
The so-called plane waves
$$ \Phi = A\exp (ip_\mu x^\mu) $$
are the simplest solutions to the Klein-Gordon equation. The momentum must obey $p_\mu p^\mu =m_0^2$. In the rest frame, $\Phi$ is independent of space but it depends on time via
$$\Phi = A\exp(-i m_0 c^2 t / \hbar) $$
where I restored the $c$ and $\hbar$ i.e. the usual units. But the angular frequency of this associated wave $\Phi$ is related to the energy via $E=\hbar\omega$, a basic formula of the quantum theory.
So the mass $m_0$ is measured as $E/c^2$ in the rest frame. The energy is calculated from the angular frequency $E=\hbar\omega$. And the angular frequency may be adjusted by adjusting the mass term $-m_0^2/2 \Phi^2$ in the Lagrangian. That's why this mass term controls the usual inertial mass.
For fermions, the box is replaced by the Dirac operator and the second power of $m_0$ becomes the first power. Different particle species have different values of the mass. Anna's answer is conveniently complementary and addresses this part of the question.
At any rate, the mass is in no way "redefined" in particle physics. It exactly agrees with everything we expected from the "mass" in the past. The equations needed to describe the creation and annihilation of elementary particles are given by QFT and QFT has to change the value of the masses – and it's still the same concept of "masses" – through the "mass terms".
