Section and pull-back bundle I find in general that the pull-back of a section of a vector bundle is a section of the pull-back bundle, but this seems to be false for the cotangent bundle. Let $\phi:M\to N$ be a smooth map, $\omega\in \Gamma(T^*N)$ be a $1$-form and $X\in \Gamma(TM)$ be a vector field. Then, since $(\phi^*\omega)(X)=\omega(\phi_*X)$, we can say that $\phi^*\omega$ is a $1$-form of $T^*M\ncong\phi^*T^*N$, in particular the fibers of $T^*N$ and $T^*M$ are different and thus the pull-back bundle $\phi^*T^*N$ cannot be equivalent to $T^*M$. $\textbf{My question}$: why?

You are confused by two different notions of pullback. The pullback of $f:M\rightarrow N$ by $g:P\rightarrow N$ is $(x,y)\in M\times P$ such that $f(x)=g(x)$.

A $1$-form defined on $N$ is a morphism $\omega:N\rightarrow T^*N$, if you have $\phi:M\rightarrow N$, you cannot defined the pullback of $\omega$ in the previous sense since the target of $\phi$ is not $T^*N$, there is another notion of pullback of a $1$-form derfined on $M$ by $(\phi^*\omega)_x(u)=\omega(d\phi_x(u))$.