Dual Transformation 
If $f : V \to W$ is a linear transformation,  a linear transformation $f^* : L^k(W) \to L^k(V)$ is defined by  $$f^*T(v_1, \dots, v_k) = T(f(v_1), \dots, f(v_k))$$ for $T \in L^k(W)$ and $v_1 \dots v_k \in V.$
Note that $L^k(W)$ denotes the set of all $k$ tensors on $W$ and we define $L^k(V)$ similiarly.

So to my understanding $f^*$ takes elements from $W^*$, the dual space of $W$ and sends those dements to the dual space of $V^*$. Now the elements of $W^*$ are linear functionals on $W$. So if $T$ really is a linear functional on $W$, why are we inputting a tuple of vectors from $V$ instead on the LHS? Also the output should be an element of $V^*$, yet the RHS is a linear functional on $W$ (because $T \in W^*_k$).
 A: Your notation is not quite right $f^*(T)$ is an element of $V^*$ it acts on a vector $v \in V$ as follows: first it sends $v$ to $W$ via $f$ to get $f(v)$ and then it evaluates $T$ at this vector. So the defining equation is 
$$f^*(T)(v)=T(f(v)).$$
A: From your notation you may be considering the dual of a multilinear , not just linear map. The map $$f^*T(v_1, \dots, v_k) = T(f(v_1), \dots, f(v_k))$$ you described is multilinear (linear in each component $v_i$ in $V_i$ ), and it seems to be defined as the dual of a multilinear map $$f: V\times V \times....\times V  \rightarrow W\times W....\times W$$ (  k times in both cases). Notice that this map is defined on the dual of $V^k$ because it takes $(v_1,v_2,..,v_k)$ as inputs. 
 The case $k=1$ is described above by Rene; notice how the map takes $v$ as an argument, so that it is an element of $V^*$. As an example of this induced map you have the case of the pullback of a differential k-form by a(n) EDIT linear map. This map is often the tangent map between the respective tangent spaces. Notice that both maps are contravariant, and that this induced map is a (contravariant, of course) functor between vector spaces .
