 Solution

Using ${\mu }^2=\lambda v^2$, expand the potential around the chosen vacuum:The linear term vanishes at the minimum. With canonical real-field kinetic terms, comparison with $m_h^2{h}^2/2+m_{\pi }^2{\pi }^2/2$ givesThe massive radial scalar changes the magnitude of the vacuum field; the massless Goldstone boson changes its orientation. The symmetry is global here, so no gauge field eats that mode.

The genuinely new scalar cubic interactions are $-\lambda v{h}^3-\lambda vh{\pi }^2$. Including identical-field factorials, their Feynman vertex factors areThe quartic interactions remain $-\lambda h^4/4-\lambda h^2{\pi }^2/2-\lambda {\pi }^4/4$, with respective vertex factors $-6i\lambda ,-2i\lambda ,-6i\lambda$. The shifted Yukawa terms from iv give $h\bar{\psi }\psi$ vertex $-ig$ and $\pi \bar{\psi }\psi$ vertex $g{\gamma }_5$. These Yukawa structures were present before the shift; the new fermion bilinear is the mass term, not an additional independent interaction coupling. There are no derivative interactions from the linear Cartesian field shift.