Show $f^*dx_i = \sum_{j=1}^l \frac{\partial f_i}{\partial y_j} dy_j = df_i$ Guillemin and Pollack's Differential Topology Page 164:
$U \subset \mathbb{R}^k$ and $V \subset \mathbb{R}^l$ be open subsets. Let $f: V \to U$ to smooth. Use $x_1, \dots, x_k$ for the standard coordinate functions on $\mathbb{R}^k$ and $y_1, \dots, y_l$ on $\mathbb{R}^l$. Write $f = (f_1, \dots, f_k)$, each $f_i$ being a smooth function on $V$. The derivative $df_y$ at point $y \in V$ is represented by the matrix $$\frac{\partial f_i}{\partial y_j}(y),$$
and its transpose map $df_y^*$ is represented by the transpose matrix. Consequently,

$$f^*dx_i  = \sum_{j=1}^l \frac{\partial f_i}{\partial y_j} dy_j = df_i.$$


My solution:
For a smooth function $f$, $df$ is linear. And $df^*$ is the adjoint of the map $df_*$.
Consider a tangent vector $Y \in T_yU$ such that
$$Y = \sum_{j = 1}^l Y^j\frac{\partial}{\partial y^j}.$$
Then we have
$$(f^*dx_i)(Y) = (f^*dx_i)\sum_{j = 1}^l Y^j\frac{\partial}{\partial y^j}.$$
$f^*$ is linear, $dx_i$ is linear, and the composition of linear function is linear. Hence $f^*dx_i$ is linear. So
$$(f^*dx_i)\left(Y^1\frac{\partial}{\partial y^1} + \cdots + Y^l\frac{\partial}{\partial y^l}\right) = (f^*dx_i)\left(Y^1\frac{\partial}{\partial y^1}\right) + \cdots + (f^*dx_i)\left(Y^l\frac{\partial}{\partial y^l}\right).$$
By Commutativity of $Y^j$.:
$$ Y^1(f^*dx_i)\left(\frac{\partial}{\partial y^1}\right) + \cdots + Y^l(f^*dx_i)\left(\frac{\partial}{\partial y^l}\right) = \sum_{j=1}^l Y^j (f^*dx_i)\left(\frac{\partial}{\partial y^j}\right).$$
Use the definition $$f^*\omega = \omega \circ f_*.$$
We have
$$\sum_{j=1}^l Y^j (f^*dx_i)(\frac{\partial}{\partial y^j}) = \sum_{j=1}^l Y^j dx_i(f_*(\frac{\partial}{\partial y^j})).$$
Because $f$ maps from $V \subset \mathbb{R}^l$ to  $U \subset \mathbb{R}^k$, it can be written as
$$f = (f_1, f_2, . . . f_k),$$
with each $f_i$ being a function of the $y_j \in V \subset \mathbb{R}^l$. Consider and $g:U \to R$ sufficiently differentiable, then the vector field $f_*(\frac{\partial}{\partial y^j})$ on $U$ may be applied to $g$:
$$f_*(\frac{\partial}{\partial y^j})[g(x_1, x_2, . . . x_k)] = \frac{\partial}{\partial y^j}(g(f_1(y_1, . . . y_l), f_2(y_1, . . . y_l), . . . f_k(y_1, y_2, . . . , y_l))),$$
according to the definition
$$(F_*X)(f) = X(f \circ F).$$
Hence,
$$\frac{\partial}{\partial y^j}(g(f_1(y_1, . . . y_l), f_2(y_1, . . . y_l), . . . f_k(y_1, y_2, . . . , y_l))) = \frac{\partial}{\partial y^j}(g(x_1, \dots, x_k)).$$
Following Chain rule,
$$\frac{\partial}{\partial y^j}(g(x_1, \dots, x_k)) = \frac{\partial g}{\partial x^1} \frac{\partial x^1}{\partial y^j} + \cdots + \frac{\partial g}{\partial x^k} \frac{\partial x^k}{\partial y^j} = \sum_{n = 1}^k \frac{\partial g}{\partial x_n} \frac{\partial f_n}{\partial y_j}.$$
That is
$$f_*\left(\frac{\partial}{\partial y^j}\right)[g(x_1, x_2, . . . x_k)] = \sum_{n = 1}^k \frac{\partial g}{\partial x_n} \frac{\partial f_n}{\partial y_j}.$$
By commutativity of first-order derivative,
$$\sum_{n = 1}^k \frac{\partial g}{\partial x_n} \frac{\partial f_n}{\partial y_j} =
\sum_{n = 1}^k \frac{\partial f_n}{\partial y_j} \frac{\partial g}{\partial x_n} .$$
Thus we see that the vector field $f_*\left(\frac{\partial}{\partial y^j}\right)$ satisfies
$$f_*\left(\frac{\partial}{\partial y^j}\right) =  \sum_{n = 1}^k \frac{\partial f_n}{\partial y_j}\frac{\partial}{\partial x_n}.$$
So
$$\sum_{j = 1}^lY^jdx_i(f_*(\frac{\partial}{\partial y^j})) = \sum_{j = 1}^lY^jdx_i\left(\sum_{n = 1}^k \frac{\partial f_n}{\partial y_j}\frac{\partial}{\partial x_n}\right).$$
As before, we use the fact that $dx_i$ is linear, and first-order derivative is commutative,
$$\sum_{j = 1}^lY^jdx_i\left(\sum_{n = 1}^k \frac{\partial f_n}{\partial y_j}\frac{\partial}{\partial x_n}\right) 
= \sum_{j = 1}^lY^j\left(\sum_{n = 1}^k dx_i \frac{\partial}{\partial x_n}\frac{\partial f_n}{\partial y_j}\right).$$
Using $dx_i(\frac{\partial}{\partial x_n}) = \delta_{in}$,
$$\sum_{j = 1}^lY^j\left(\sum_{n = 1}^k dx_i \frac{\partial}{\partial x_n}\frac{\partial f_n}{\partial y_j}\right) 
= \sum_{j = 1}^lY^j\left(\frac{\partial f_i}{\partial y_j}\right)
= \sum_{j = 1}^l (dy_j Y)\left(\frac{\partial f_i}{\partial y_j}\right).$$
So,
$$\sum_{j = 1}^l (dy_j Y)\left(\frac{\partial f_i}{\partial y_j}\right)
=\sum_{j = 1}^l\left(\frac{\partial f_i}{\partial y_j}\right) (dy_j Y).$$
According to Show that $d\phi = \sum \frac{\partial \phi}{\partial x_i}dx_i.$ We have
$$df = \sum \frac{\partial f}{\partial y_i}dy_i.$$
So
$$\sum_{j = 1}^l\left(\frac{\partial f_i}{\partial y_j}\right) (dy_j Y)
=df_i (Y).$$
Since this holds for any $Y \in T_yV$, we have shown that
$$f^*dx_i = \sum_{j = 1}^l \frac{\partial f_i}{\partial y_j}dy_j = df_i$$
 A: I'd like to point out, before trying to answer this question, that your definitions confuse me a little.  For instance, what is $I(y)$?  And in your expression for $\omega$, 
$\omega = \sum_{1 \le i_1 < . . . i_k \le n}I(y)dx_i$,
why is there apparently a multi-index of some sort on the $\Sigma$ symbol which doesn't seem (to me at least) to occur in the summand $I(y)dx_i$?  Well, perhaps this stuff is explained in Guillemin and Pollack, which I haven't looked at in quite awhile, fine book though it be.  Not trying to be critical here, merely seeking clarification.
Having said these things, let's try to prove that
$f^*dx_i = \sum_{j = 1}^l\frac{\partial f_i}{\partial y_j}dy_j = df_i$.
I'm going to try to do this the way I learned it, mostly from general principles; as I indicated above, I don't have a copy of Guillemin and Pollack in front of me, so if I seem like I'm winging it, bear with me . . .
The first thing you need to know is that $f^*$ is the adjoint of the map $f_*$, in the sense usually used in linear algebra:  if $T:V \to W$ is a linear map between vector spaces $V$ and $W$, then for any $\sigma \in W^*$ we define the linear functional $T^*\sigma \in V^*$ by the formula $T^*\sigma(v) = \sigma(Tv)$ for vectors $v \in V$.  Thus it is easily seen that $T^*:W^* \to V^*$ is also a linear map.  This idea of course applies pointwise with respect to $U$ and $V$, i.e. fiberwise with respect to the tangent and cotangent spaces of $U$ and $V$.  Now let's look at $f^*dx_i$.  For any tangent vector
$Y \in T_yU$, we have
$(f^*dx_i)(Y) = dx_i(f_*(Y))$,
and with
$Y = \sum_{j = 1}^l Y^j\frac{\partial}{\partial y^j}$
we obtain
$(f^*dx_i)(Y) = (f^*dx_i)(\sum_{j = 1}^l Y^j\frac{\partial}{\partial y^j})$,
and by linearity of everything this yields
$(f^*dx_i)(Y) = \sum_{j = 1}^lY^j (f^*dx_i)(\frac{\partial}{\partial y^j}) =  \sum_{j = 1}^lY^jdx_i(f_*(\frac{\partial}{\partial y^j}))$.
We scrutinize $f_*(\frac{\partial}{\partial y^j})$.  With $f = (f_1, f_2, . . . f_k)$, with each $f_p$, $1 \le p \le k$ being a function of the $y_q$, $1 \le q \le l$, and $g:U \to R$ and sufficiently differentiable, the vector field $f_*(\frac{\partial}{\partial y^j})$ on $U$ may be applied to $g$:
$f_*(\frac{\partial}{\partial y^j})[g(x_1, x_2, . . . x_k)] = \frac{\partial}{\partial y^j}(g(f_1(y_1, . . . y_l), f_2(y_1, . . . y_l), . . . f_k(y_1, y_2, . . . , y_l)))$
$= \sum_{n = 1}^k \frac{\partial g}{\partial x_n} \frac{\partial f_n}{\partial y_j}$,
this last equality following from the fact that $x_p = f_p(y_1, y_2, . . . , y_l)$ and the chain rule.  Thus we see that the vector field $f_*(\frac{\partial}{\partial y^j})$ satisfies
$f_*(\frac{\partial}{\partial y^j}) =  \sum_{n = 1}^k \frac{\partial f_n}{\partial y_j}\frac{\partial}{\partial x_n} $,
and if this is inserted into our previous expression for $(f^*dx_i)(Y)$,
$(f^*dx_i)(Y) = \sum_{j = 1}^lY^jdx_i(f_*(\frac{\partial}{\partial y^j}))$,
it follows that
$(f^*dx_i)(Y) = \sum_{j = 1}^l \sum_{n = 1}^k Y^j dx_i(\frac{\partial f_n}{\partial y_j}\frac{\partial}{\partial x_n})$,
and using $dx_i(\frac{\partial}{\partial x_n}) = \delta_{in}$,
$(f^*dx_i)(Y) = \sum_{j = 1}^l Y^j \frac{\partial f_i}{\partial y_j} =  \sum_{j = 1}^l \frac{\partial f_i}{\partial y_j}dy_j(Y) = df_i(Y)$,
since we have $Y = \sum_{j = 1}^l Y^l \frac{\partial}{\partial y_l}$ and $dy_j(Y) = Y^j$.  Since this holds for any $Y \in T_yV$, we have shown that 
$f^*dx_i = \sum_{j = 1}^l \frac{\partial f_i}{\partial y_j}dy_j = df_i$,
as per request.  QED.
Whew!  Too many indices and subscripts to keep track of!
Now, as I recall, establishing this formula on $1-$forms allows it to be extended in the usual manner to all of $\Lambda(T*U)$, i.e., all form (fields) by careful use of the definitions of the $\wedge$ product and a lot of maneuvering of matrices and indices.
Gotta run, my night job beckons.
A: We know $f^*dx_i$ is tensor field and compute it at $p\in V$ which is arbitary,$(f^*{dx_i})_p$=${dx_i}_{f(p)}{f_*}_p$(from definition of pullback),also we have ${f_*}_p=df_p=\Big[\frac{\partial f_t}{\partial y_j}(p)\Big]_{k\times l},$ and ${dx_i}_{f(p)}={{x_i}_*}_{f(p)}=[0,...,0,1,0,...,0]_{1\times k}$(with 1 in ith place),the multiplication of two matrices is $\Big[\frac{\partial f_i}{\partial y_j}(p)\Big]_{1\times l}$ which is equal to ${df_i}_p$,and ${df_i}_p=\sum_{j=1}^l \frac{\partial f_i}{\partial y_j}(p){ dy_j}_p$.
A: First, I think both answers offered thus far are excellent in their own way. I will merely attempt to say the same with a slightly different notation.
First, $f=(f^1, \dots, f^k)$ is a function of $y = (y^1, \dots , y^l)$. Therefore, we differentiate $f$ with respect to $y$ at the point $y=p$ we have: 
$$ df_p(h) = \sum_{i=1}^k\sum_{j=1}^l[f'(p)]_{ij}h_j\frac{\partial}{\partial x^i}\bigg|_{f(p)} \qquad \text{where} \ [f'(p)]_{ij} = \frac{\partial f^i}{\partial y^j} $$
where $f'(p) \in \mathbb{R}^{k \times l}$ and $[h_j] \in \mathbb{R}^l$ thus $f'(p)[h] \in \mathbb{R}^k$. I suppose, to be explicit, $h = \sum_{j=1}^l h_j\frac{\partial}{\partial y^j}\bigg|_p$. Here I, as is my custom, assume $\mathbb{R}^n = \mathbb{R}^{1 \times n}$, that is, euclidean space is made of column vectors. I'll place $dx^j$ at $f(p)$ but I'll forego adorning $dx^j$ with that point-dependence for brevity in what follows. The pull-back of $dx^j$ from the range of $f$ will form (pun-intended) a one-form $\alpha$ in the domain at $p$. We can calculate such a one-form by calculating its components in the $\{dy^1, \dots , dy^l \}$ basis for $T^*V_p$ by evaluation at the coordinate vector fields to which these are dual. Consider, 
$$ \bigl((f^*)_{p}(dx^m)\bigr)(\frac{\partial}{\partial y^n}\bigg|_p) = dx^m ( df_p(\frac{\partial}{\partial y^n}\bigg|_p)) \qquad \star$$ 
Now, $h_j = \delta_{jn}$ for the coordinate vector field ($h = \sum_{j=1}^{l}\delta_{jn}\frac{\partial}{\partial y^n}\bigg|_p)$ hence
$$ df_p(\frac{\partial}{\partial y^n}\bigg|_p)) =
\sum_{i=1}^k\sum_{j=1}^l[f'(p)]_{ij}\delta_{jn}\frac{\partial}{\partial x^i}\bigg|_{f(p)} = 
\sum_{i=1}^k[f'(p)]_{in}\frac{\partial}{\partial x^i}\bigg|_{f(p)} $$
Finally, feed this into the $\star$ equation,
\begin{align} \bigl((f^*)_{p}(dx^m)\bigr)(\frac{\partial}{\partial y^n}\bigg|_p) &= dx^m (\sum_{i=1}^k[f'(p)]_{in}\frac{\partial}{\partial x^i}\bigg|_{f(p)}) \\
&= \sum_{i=1}^k[f'(p)]_{in} dx^m (\frac{\partial}{\partial x^i}\bigg|_{f(p)}) \\
&= \sum_{i=1}^k[f'(p)]_{in}\delta_{im} \\
&= [f'(p)]_{mn}
\end{align}
Therefore,
$$ (f^*)_{p}(dx^m) = \sum_{n=1}^l [f'(p)]_{mn} dy^n = \sum_{n=1}^l \frac{\partial f^m}{\partial y^n}dy^n = df^m$$
