Definition of pull back operation 
Let $\varphi:U \rightarrow V$ be a differentiable map between open sets $U \subset \mathbb{R}^n$ and $V \subset \mathbb{R}^m$. Define the pull back operation $\varphi^*: \Omega^{k}(V) \rightarrow \Omega^k(U)$.

I said this is just
$$\varphi^*\omega = \sum_{1 \leq i_1<\cdots < i_k \leq m}c_{i_1, \cdots i_k}(f_1(x_*), \cdots f_m(x_*)) df_{i_1} \wedge \cdots \wedge df_{i_k} \in \Omega^k(U)$$
where $x_* = (x_1, \cdots x_n)$.
Is this correct?
 A: The best way is to start by the definition. Let's begin with $1$-forms so that we can understand what's the reasoning behind it. Recall that $\omega \in \Omega^1(U)$ is simply a way to associate at each point a linear function of one single vector. In other words if $p \in U$ then $\omega(p) \in T^\ast_p U$. Our problem is: $\varphi$ transforms $U$ into $V$, is there a way to produce one-forms on $U$ from those at $V$? In other words, how could we transform the forms back, pull them back. The idea is pretty straightforward: we already now that $\varphi_\ast$ (the pushforward) pushes the vectors to the corresponding tangent space at $V$, so the most possible reasonable way to define this pullback operation is by setting:
$$\varphi^\ast(\omega)(v)=\omega(\varphi_\ast(v))$$
In other words: the action of the 1-form $\varphi^\ast(\omega)$ on some vector $v \in T_pU$ is the same as the action of $\omega \in \Omega^1(V)$ on the pushed vector. Think a bit about it, there's a natural correspondence between the vectors on each tangent space, and this natural correspondence induces a natural correspondence between the 1-forms.
Now for $k$-forms. Recall that a $k$ form $\omega \in \Omega^k(U)$ is an association of an alternating $k$-tensor at each point of $U$. In that sense, for each $p \in U$, $\omega$ associates one alternating function of $k$ vectors $\omega(p)$. So, we use the same reasoning, we push the $k$ vectors to the corresponding tangent space at $V$ and apply the form there (since we know the form there). This is simply:
$$\varphi^\ast(\omega)(v_1, \dots, v_k)= \omega(\varphi_\ast(v_1), \dots, \varphi_\ast(v_k))$$
So this is the pullback you want with motivations on the way. At each point $p \in U$ the field $\varphi^\ast(\omega)$ will associate a function of $k$ vectors. We know how to apply $\omega(\varphi(p))$, so what we do? We push the vectors to $T_{\varphi(p)} V$ and and apply $\omega(\varphi(p))$ on the pushed vectors. Now you can compute it in terms of local coordinates using the expression of the pushforward in local coordinates.
From your question I didn't understand if you wanted the definition or the expression in local coordinates. By the way this is the definition. Anything more I can help you let me know. Good luck!
Obs: these are very good notes to read about it.
EDIT: About the equation provided in the question. There are a few relations that can be proved rather easily directly from the definition of the pullback made above. If $\varphi : U \to V$ is a smooth map and if $\omega, \omega_1, \omega_2  \in \Omega^k(V)$ and if $f : V \to \mathbb{R}$ then we have:


*

*$\varphi^\ast(\omega_1 + \omega_2) = \varphi^\ast(\omega_1)+\varphi^\ast(\omega_2)$

*$\varphi^\ast(f\omega) = (f\circ \varphi)\varphi^\ast(\omega)$

*$\varphi^\ast(\omega_1 \wedge \omega_2) = \varphi^\ast(\omega_1) \wedge \varphi^\ast(\omega_2)$


Also if $y^i : V \to \mathbb{R}$ are the coordinate functions on $V$, we have:


*$\varphi^\ast(dy^i) = d\varphi^i$


Now consider that the coordinate functions of $\varphi$ are $f^i$ and that we have $\omega \in \Omega^k(V)$ given by:
$$\omega = \sum_{i_1 < \cdots < i_k}{c_{i_1 \dots i_k}dy^{i_1}\wedge \cdots \wedge dy^{i_k}}$$
Apply the pullback to this thing and use properties 1 - 4, you'll get:
$$\varphi^\ast \omega = \sum_{i_1 < \cdots < i_k}{(c_{i_1 \dots i_k}\circ \varphi)df^{i_1}\wedge \cdots df^{i_k}}$$
And since the coordinate functions of $\varphi$ are $f^i$ this is rewritten as what you gave at first:
$$\varphi^\ast \omega (p) = \sum_{i_1 < \cdots < i_k}{c_{i_1 \dots i_k}(f^1(p), \dots, f^m(p))df^{i_1}\wedge \cdots \wedge df^{i_k}}$$
Recalling just that $\varphi^\ast$ is a function defined on $U$ and letting the ranges of the indices be understood in context. Also look on the properties being used: first we use 1 and put the pullback inside the sum, then we use 2 and compose the coordinate functions of $\omega$ with $\varphi$, then we use 3 to write the pullback of that whole wedge product as the wedge product of the pullbacks and finally we use 4 to write the pullback of each differential of coordinate as the differential of the corresponding coordinate of $\varphi$.
I hope this helps you out. Good luck!
