Cases where multiple induction steps are provably required

Here is a reference for one way of making precise sense of your question and answering it:

Stefan Hetzl and Tin Lok Wong (2017): "Some observations on the logical foundations of inductive theorem proving", Logical Methods in Computer Science, Volume 13, Issue 4, doi:10.23638/LMCS-13(4:10)2017, arXiv:1704.01930

In Section 2.4 they show the following. Let $\text{PA}^-$ be the theory of the non-negative parts of discretely ordered rings (language: $\langle 0,1,+,\times,<\rangle$—see the paper for an axiomatization). Then for each theorem $\sigma$ of PA there is a formula $\varphi(x)$ such that $\text{PA}^-$ proves $\varphi(0)$ and $\forall x(\varphi(x)\implies\varphi(x+1))$ and $\forall x.\varphi(x)\implies\sigma$. Thus in this sense one application of induction always suffices.

Here is an answer concerning recursion, rather than induction, but they are of course related.

Namely, the Ackermann function is defined by a double nested recursion $$A(m+1,n+1)=A(m,A(m+1,n))$$ with anchor cases defining $A(0,n)$ and $A(m,0)$. The function exhibits extremely rapid growth.

I mention the function because one can prove that the Ackermann function is not a primitive recursive function, which are the functions one can construct from some primitive functions by closing under composition and simple recursion.

Thus, the Ackermann function can be defined by a nested recursion, but not by a simple recursion.