The mass of the Higgs particle is conceptually independent from the mass Yukawa couplings impart to fermions. If you opened a SM QFT book or review, you'd find an illustration of spontaneous symmetry breaking summary, illustrated by the abelian version of the linear σ model of Gell-Mann and Levy (1964),
whose relevant terms in the Lagrangian are
$$
...+ y\bar \psi (\sigma +i\pi \gamma_5)\psi+ \frac{\lambda}{4} (\sigma^2 + \pi^2 -v^2)^2 +...
$$
The σ is the analog of the Higgs particle and the π the analog of the Nambu-Goldstone boson.
By contrivance, at the bottom of the ("sombrero") potential, it happens that $\langle \sigma \rangle =v$ and $\langle \pi\rangle=0$. Define $\sigma = h+v$, so the above chunk of the Lagrangian reads
$$
y\bar \psi (v+ h+i\pi \gamma_5)\psi+ \frac{\lambda}{4} ( h^2+ 2v h \pi^2)^2\\
= yv\bar \psi \psi+ {\lambda} v^2 h^2+ ... ~,
$$
where only terms bilinear in the fields are kept.
Note there is no mass term for the goldston, 𝜋, and masses for the fermion and the Higgs,
$$
m_\psi= - yv, \qquad m_h^2=2\lambda v^2.
$$
They have nothing to do with each other. They are related to v, the location of the bottom of the potential, but the relevant dimensionless couplings multiplying them are y and 𝜆, respectively, completely, and mysteriously, unrelated. The Higgs mass determines its fluctuations around the bottom of the potential, while the goldston is massless and "fluctuates" with enormous/infinite wavelength into and out of the vacuum.
But it's all math: you can't spin qualitative tales on math, unless you are blasé about the math underlying it...
