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How the field and interactions are changed when we assume that proton has finite radius in atom for example? What gives the finite size effect? Is it the higher moments of multipole expansion?

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That depends on the energy at which you probe the system. At low energy nothing changes, at high energy it stops being sensible to talk about the proton any more. In between there is a difficult region where things are strongly changed, but the proton is still a good unit of interaction at least part of the time. – dmckee♦ Sep 18 '12 at 17:06
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One may also solve the corrections to the energy levels of the electron coming from the nonzero proton radius. It's a nice homework exercise in a good enough undergraduate course. There is a correction in the first order of perturbation theory, from the expectation value of the $\Delta H$ in the ground states. $\Delta H$ corresponds to a modified potential - pointlike protons have $1/r$ everywhere but nonzero-size protons have the potential plateaued inside a ball of radius $R$. One needs to take a difference. It will only markedly influence the states with $l=0$ whose $\psi(r=0)\neq 0$. – Luboš Motl Sep 18 '12 at 18:11

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A finite size of the proton induces electromagnetic form factors in the QED interactions and makes them nonlocal.

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