Struck in Diffeomorphism invariance of the line integral I have been trying to study The Cotangent bundle from An introduction to smooth manifolds by John M Lee. I have been struck at a specific point in the Diffeomorphism Invariance of the integral. The proposition is given below.
Let $\omega$ be a smooth covector field on the compact interval $[a,b] \subseteq \mathbb{R} $. If $\varphi : [c,d] \to [a,b]$ is an increasing diffeomorphism, then
$$
\int_c^d \varphi ^ * \omega = \int_a^b \omega
$$
The proof starts by giving the coordinate expression for $\varphi ^ * \omega$ as
$$
(\varphi ^ * \omega)_s = \mathit{f}(\varphi(\mathit(s))\varphi'(\mathit{s})ds
$$
Where $ \omega(t)=f(t)\,dt$ for some smooth function $ f:[a,b]\to \mathbb{R}$, and $\mathit{t}$ is the standard coordinate on $[a,b]$, and $\mathit{s}$ is the standard coordinate on $[c,d]$.
I have been able to work out that 
$$
(\varphi ^ * \omega)=(\mathit{f} \circ \varphi) d(t \circ \varphi)
$$
From this I am unable to reach the coordinate expression for $(\varphi ^ * \omega)_s$. Please help me by showing how the coordinate expression is arrived at.

The expression $f \circ \varphi$ is literally the same as the function $f(\varphi(s))$.
Now for the second part: the $d(t \circ \varphi)$. The function $t \circ \varphi$ means apply the function $\varphi$, and then take the $t$-coordinate of the result. Since everything is 1-dimensional, "taking the $t$-coordinate" is kind of silly. It's kind of assumed from your notation that $\varphi$ takes the $s$-coordinate to the $t$-coordinate, so $t \circ \varphi$ might as well just be written as $\varphi$, and the meaning would be the same.
So $d(t \circ \varphi)$ is pretty much just $d\varphi$, which is $\varphi'(s) ds$.
