I understand that in beta decay, a quark changes its flavor and emits a $W$-Boson, and this $W$ (is the W Boson virtual, by the way?) quickly decays into an electron and an electron antineutrino or positron and neutrino. What makes the $W$-Boson a gauge boson? I don't see any fundamental force being exchanged, and the W seems to act as an intermediate particle.
Processes in elementary particle physics are described by Quantum Field Theory (QFT). In this theory the fields (eletromagnetic force, weak force etc.) are quantized, so they are presented by particles. The attraction, but also repulsion between particles, even decays are represented by the exchange of a gauge boson. Physicists use for this purpose the famous Feynman diagrams (they are very suggestive diagrams, but actually one has to be careful about the interpretation of these diagrams, each part of such a diagram corresponds to mostly complicated mathematical expressions). So the attraction respectively the repulsion of 2 charged particles is represented by the exchange of photons, and the beta decay is represented by the exchange of another boson, the W-boson.
They are called gauge-bosons since their corresponding force respects some local symmetry --- actually it is a freedom (which is called gauge freedom) in the choice of the corresponding potentials of the force -- in case of a photon it is called U(1), and in case of the W-boson it is SU(2).
But, I think the main message to keep here is that forces between elementary particles are described in QFT by the exchange of a particle, actually always bosons (excluding here SUSY). One can make a little computational exercise (you can look it up in the book of Zee, "QFT in a nutshell" in the first chapter) and show that this approach (exchange of a particle) leads to a decrease (increase in case of repulsion) of the energy of in the interaction participating particles, which is nothing else as a felt force.