How can you express $f^*\left(\sum_{j_1,\dots,j_k} a_{j_1\dots j_k}dy^{j_1}\otimes\cdots\otimes dy^{j_k}\right)$ in terms of $dx^i$? Suppose $f\colon M^n\to N^m$ is a map between manifolds, with $(x,U)$ and $(y,V)$ coordinates systems around $p$ and $f(p)$. 
How can you express $f^*\left(\sum_{j_1,\dots,j_k} a_{j_1\dots j_k}dy^{j_1}\otimes\cdots\otimes dy^{j_k}\right)$ in terms of $dx^i$?
I've managed to derive the formula 
$$
(f^* dy^j)(p)=\sum_{i=1}^n\frac{\partial (y^j\circ f)}{\partial x^i}(p)\cdot dx^i(p)
$$
which seems pertinent in rewriting the $dy^{j_k}$ in terms of $dx^i$, but I don't know how $f^*$ acts on tensors. This comes from Chapter 4 of Michael Spivak's A comprehensive Introduction to Differential Geometry, I've been scouring the chapter but am unsure of what $f^*$ does. Can anyone explain it? Thanks.
 A: You can derive the action of the pull-back quickly in terms of the push-forward. Assume that you have a vector $X\in TM$. Then $f_*X \in TN$ and (assuming $a$ is a one-form) you can consider the standard pairing of vectors and convectors. By definition, this satisfies $\langle f_* X, a \rangle = \langle X, f^*a\rangle$.
To make things explicit, assume that $X$ is generated by some curve $\gamma(t):\mathbb{R} \to M$. In coordinates at some $x \in X$, $X^m = \frac{d\gamma^m}{dt}\vert_{t=0}$. The push forward to $f(x) \in N$ is then simply defined by the curve $f \circ \gamma:\mathbb{R} \to N$, so partial differentiation gives $(f_*X)^m = \sum \frac{f^m}{\partial x^n} X^n$. Then $\langle f_* X, a\rangle = \sum \frac{\partial f^m}{x^n} X^n a_m$, and it follows that $f^*a = \sum (f^*a)_n dx^n = \sum \frac{f^m}{x^n} a_m dx^n$.
In principle you need to garnish the above expressions with where you're evaluating everything, but that can get really confusing if you try to do it in the notation. Simply add $\vert_x$ to all expressions on $M$ and $\vert_{f(x)}$ to all those in $N$.
If you're struggling with the derivation for higher degree forms, consider that for a $p$-form $a$ on $N$ and $p$ vectors $X^{(p)}$ on $M$ a similar relation holds: $a(f_* X^{(1)}, \dots, f_* X^{(p)}) = (f^*a)(X^{(1)}, \dots, X^{(p)})$.
