Let $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a smooth transformation. Define the pullback $T^*: C^k (\mathbb{R}^m) \rightarrow C^k (\mathbb{R}^n)$ (With $C^k(\mathbb{R}^n)$ being the set of functionals on $k$-cells on $\mathbb{R}^n$) by $T^*: Y \mapsto Y \circ T$.
Thus, the pullback of $Y \in C^k (\mathbb{R}^m)$ is the functional on $C_k (\mathbb{R}^n)$ $$T^*Y:\phi \mapsto Y(T\circ \phi)$$
Why is the pullback linear, and why does $(T \circ S)^* = S^* \circ T^*$?
Let $X,Y$ be arbitrary functionals (and $a$ an arbitrary constant). Since linearity is preserved on $k$-forms:
\begin{align} T^*(aY + X) &= (aY + X) \circ T \\ &= (aY + X) (T) \\ &= aY(T) + X(T) \\ &= aT^*(Y) + T^*(X). \end{align}
\begin{align} (T \circ S)^* (Y) &= Y \circ (T \circ S) \\ &= Y \circ T \circ S \\ &= S^*(Y \circ T) \\ &= S^* \circ T^* (Y). \end{align}