Differential as bundle map and pull back bundle In the wikipedia article about the pushforward, it is stated that if $f: M\to N$ is smooth, then it induces a bundle map $df: TM \to TN$. It is then claimed that equivalently $f_*=df$ is a bundle map from $TM$ to the pullback bundle $f^* TN$.
Why is this equivalent? The bundles $TN$ and $f^* TN$ are clearly not the same as they are bundles over different spaces. They could be isomorphic, but is still seems strange that this would hold independently of $f$.
Edit: The full quote is

Equivalently (see bundle map), φ∗ = dφ is a bundle map from TM to the pullback bundle φ∗TN over M, which may in turn be viewed as a section of the vector bundle Hom(TM, φ∗TN) over M. The bundle map dφ is also denoted by Tφ and called the tangent map. In this way, T is a functor. 

 A: Recall the definition of the pull-back bundle (in the context you are interested in): If $f: M\to N$ is a smooth map, $df: TM\to TN$ is the differential of $f$, then (as a topological space) 
$$
f^*(TN)=\{(x,\xi): x\in M, \xi\in TN, \xi\in T_{f(x)}N\}\subset M\times TN. 
$$
The bundle structure on $f^*(TN)$ is given by the projection 
$$
\pi: (x,\xi)\mapsto x. 
$$
Now, $df$ defines a bundle map $TM\to f^*(TN)$ which I will denote $Df$:
$$
Df: (x,\eta)\mapsto (x, df_x(\eta)), x\in M, \eta\in T_xM.
$$
I will leave it to you to check that $Df$ is indeed a morphism of vector bundles 
$$
Df: TM\to f^*(TN). 
$$
By abuse of notation, one frequently denotes the morphism $Df$ simply $df$ since the"interesting" parts of these morphisms are the same. 
Given a vector field $X\in {\mathfrak X}(M)$, the push-forward $f_*(X)$ is the section $Y$ of $f^*(TN)$ which is given by the formula
$$
Y_x= Df(X_x). 
$$
That's all what there is to it. The wikipedia article commits a minor and common abuse of notation by using the notation $df$ for $Df$. 
