Let $M$ and $N$ be two smooth manifolds homeomorphic to $S^n$ such that $N=\partial B$, for some (n+1)-smooth manifold $B$.
Suppose that there is a diffeomorphism $f:M \rightarrow N$. Can I find a manifold $C$ such that $M= \partial C$ and $C$ is diffeomorphic to $B$? In other words, I want to "pullback" $B$ using $f$ (or an extension) to obtain such a manifold $C$.
(This is not true in the case that $f$ is a homeomorphism)
Thanks in advance.
Sure; you just take $B$ and "replace" each point of $N$ with the corresponding point of $M$. There is nothing deep here at all; this is just the usual argument that isomorphic structures are interchangeable.
Explicitly, let us first assume that the set $B$ is disjoint from $M$. In that case, define the set $C$ to be $(B\setminus N)\cup M$, and define $g:C\to B$ by $g(c)=c$ if $c\in B\setminus N$ and $g(c)=f(c)$ if $c\in M$. There is a unique way to make $C$ a smooth manifold with boundary such that $g$ is a diffeomorphism (namely, take the topology to consist of all sets $g^{-1}(U)$ for $U\subseteq B$ open, and define the charts on $C$ to be the composition of $g$ with charts on $B$). Since $g$ is a diffeomorphism, it must restrict to a diffeomorphism $\partial C\to \partial B$. Since $\partial B=N$ and $g^{-1}(N)=M$, we have $\partial C=M$ as a set. Moreover, the restriction of $g$ to $M$ is just the map $f$, which we assumed was a diffeomorphism for the given smooth manifold structure on $M$. So $f$ is a diffeomorphism $M\to N$ both for the given smooth manifold structure on $M$ and for the smooth manifold structure obtained by considering $M$ as the boundary of $C$. Thus these two smooth manifold structures on $M$ are the same (the identity map is a diffeomorphism between them). That is $\partial C$ is $M$ with its original smooth manifold structure.
In case $B$ is not disjoint from $M$, you just have to make it disjoint first. Take $B'$ to be some set with a bijection $h:B\to B'$ such that $B'\cap M=\emptyset$. Make $B'$ a smooth manifold in the unique way such that $h$ is a diffeomorphism. Let $N'=\partial B'$ and let $f':M\to N'$ be the composition $h|_N\circ f$, which is a diffeomorphism since $h|_N$ and $f$ are. We can now apply the previous paragraph with $(B',f',N')$ in place of $(B,f,N)$.