It's more of a mathematical tool rather than some physical interaction. To see what the maths is, we try and use the Higgs mechanism on a very simple case, which will be an abelian $U(1)$ gauge theory, and you will in the end see where the mass comes from.
The $U(1)$ invariant kinetic term of the photon is: $$\mathcal{L}_{kin}=-\frac14F_{\mu\nu}F^{\mu\nu}$$ where $$F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\ .$$
That is, $\mathcal{L}_{kin}$ is invariant under the transformation $A_{\mu}(x)\to A_{\mu}(x)-\delta_{\mu}\eta(x)$ for any $\eta$ and $x$.
Now, if we try to naively add a mass term for the photon:
$$\mathcal{L}=-\frac14F_{\mu\nu}F^{\mu\nu}+\frac12m^2A_{\mu}A^{\mu}$$
we soon find out that the mass terms violates local gauge symmetry, and hence the $U(1)$ gauge symmetry thus requires the photon to be massless.
But what happens if we can break the symmetry?
We try to do this by introducing a complex scalar field with charge $-e$ that couples to the photon as well as with itself:
$$\mathcal{L}=-\frac14F_{\mu\nu}F^{\mu\nu}+(D_{\mu}\phi)^{\dagger}(D^{\mu}\phi)-V(\phi)$$ where $D_{\mu}=\partial_{\mu}-ieA_{\mu}$ and $V(\phi)=-\mu^2\phi^{\dagger}\phi+\lambda(\phi^{\dagger}\phi)^2$. We can see that the Lagrangian is invariant under the gauge transformations:
$$A_{\mu}(x)\to A_{\mu}(x)-\partial_{\mu}\eta(x)$$
$$\phi(x)\to e^{ie\eta(x)}\phi(x)\ .$$
If $\mu^2<0$, the state of minimum energy will be that with $\phi=0$ and the potential will preserve the symmetries of the Lagrangian. Then the theory is simply normal QED with an extra charged scalar field $\phi$ with mass $\mu$.
However, if $\mu^2<0$, the field $\phi$ will acquire a vacuum expectation value:
$$\langle \phi \rangle =\sqrt{\frac{\mu^2}{2\lambda}}\equiv \frac{v}{\sqrt{2}}$$ and the global $U(1)$ symmetry will be spontaneously broken!
We can parametrize $\phi$ as:
$$\phi=\frac{v+h}{\sqrt{2}}e^{i\frac{\chi}{v}}$$
where $h$ and $\chi$ are referred to as the Higgs boson and the Goldstone boson respectively. They are real scalar fields with no vacuum expectation values. Substituting, we find:
$$\begin{align*}\mathcal{L}=&-\frac14F_{\mu\nu}F^{\mu\nu}-evA_{\mu}\partial^{\mu}\chi\\&+\frac{e^2v^2}{2}A_{\mu}A^{\mu}+\frac12(\partial_{\mu}h\partial^{\mu}h-2\mu^2h^2)\\&+\frac12\partial_{\mu}\chi\partial^{\mu}\chi+\dots\end{align*}$$
This now describes a theory with a massive photon with mass $m_A=ev$, a Higgs boson $h$ with $m_h=\sqrt2\mu=\sqrt{2\lambda}v$ and a massless Goldstone $\chi$. We can remove the Goldstone boson from the theory with a transformation called the unitary gauge, but that's beside the point.
Thus we have successfully incorporated mass into our gauge boson with the help of symmetry breaking using the Higgs mechanism.
Although this does not happen in our universe, what (probably) does happen is that the gauge symmetry of the electroweak force $SU(2)\times U(1)$ is spontaneously broken to give the gauge bosons of the weak force their mass (the photons remains massless due to $SU(2)_L\times U(1)_Y\to U(1)_Q$, i.e. electromagnetism is unbroken by the scalar vacuum expectation value). The fermions in an analogous (but non-trivial) fashion gets their mass from the mechanism.
You can see that nowhere above have we mentioned 'interaction' because the Higgs mechanism is not an interaction (although the public eats such words easily). The correct interpretation of an 'interaction' is as Anna has mentioned in her answer, so I won't elaborate on that.