A couple of things in the following examples that are not quite clear to me. I'll write the full proof and my question in the middle.
Let $S^1$ be the unit circle in the $xy$ plane. By proposition 24.1, because $S^1$ is connected, $H^0(S^1) = \mathbb{R}$ and because $S^1$ is one dimensional we have $H^k(S^1) = 0$ for all $k \geq 2$. It remains to compute $H^1(S^1)$.
Question Before I go any further with the proof: In the next part of the proof am I right that the author is trying to prove an isomorphism between $H^1(S^1)$ and $\mathbb{R}$? More specifically he will be explicitly constructing such isomorphism.
Recall from subsection 18.7 that the map $h : \mathbb{R} \to S^1, h(t) = (\cos(t),\sin(t))$. Let $i : [0,2\pi] \to \mathbb{R}$ be the inclusion map. Restricting the domain of $h$ to $[0,2\pi]$ gives us a parametrization $F := h \circ i : [0,2\pi] \to S^1$ of the circle. In Example 17.15 and 17.16 we found a nowhere-vanishing 1-form $\omega = -ydx + xdy$ on $S^1$ and showed that $F^* \omega = i^*h^*\omega = i^* dt = dt$. Thus, $$ \int_{S^1} \omega = \int_{F[0,2\pi]} \omega = \int_{[0,2\pi]} F^* \omega = \int_0^{2\pi} dt = 2\pi. $$ Since the circle has dimension 1, all the 1 forms on $S^1$ are closed, so $\Omega^1(S^1) = Z^1(S^1)$. The integration of a 1 form on $S^1$ defines a linear map $$ \varphi : Z^1(S^1) = \Omega^1(S^1) \to \mathbb{R}, \;\; \varphi(\alpha) = \int_{S^1} \alpha. $$ Because $\varphi(\omega) = 2\pi \neq 0$, the linear map $\varphi : \Omega^1(S^1) \to \mathbb{R}$ is onto.
Question I am assuming this is true since for every $\gamma \in \mathbb{R}$ we have $\varphi\left(\frac{\gamma}{2\pi}\omega\right) = \gamma$, correct? So that the equation $\varphi(x) = \gamma$ has solution.
By Stokes's theorem, the exact 1-form on $S^1$ are in $\text{ker} \varphi$.
Question Let $\omega \in \Omega^1(S^1)$ such that we have $\tau \in \Omega^0(S^1)$ with $\omega = d\tau$ we then have
$$ \int_{S^1} \omega = \underbrace{\int_{S^1} d\tau = \int_{\partial S^1} \tau}_{\text{Stokes's Theorem}} = \int_{\emptyset} \tau = 0 $$
Conversely we will show that all the 1-forms in $\text{ker}\varphi$ are exact. Suppose $\alpha = f\omega$ is a smooth 1-form on $S^1$ such that $\varphi(\alpha) = 0$.
Question Does it author mean that if $\alpha \in \Omega^1(S^1)$ then we can always express it as $\alpha = f \omega$? To me this follows from
$$ \alpha = f \omega \iff \left\langle \alpha , -y \frac{\partial}{\partial x} + x \frac{\partial}{\partial y} \right\rangle = \underbrace{(x^2 + y^2)}_{=1} f = f $$
Is this reasoning correct?
Let $\overline{f} = h^* f = f \circ h \in \omega^0(\mathbb{R})$. Then $\overline{f}$ is periodic of period $2\pi$ and $$ 0 = \int_{S^1} \alpha = \int_{F[0,2\pi]} \alpha = \int_0^{2\pi} F^* \alpha = \int_0^{2\pi} (i^*h^*f)(t) F^* \omega = \int_0^{2\pi} \overline{f}(t)dt. $$
The proof continues by first proving the following Lemma
Lemma 24.5 Suppose $\overline{f}$ is a $C^{\infty}$ periodic function of period $2\pi$ on $\mathbb{R}$ on $\mathbb{R}$ and $\int_0^{2\pi} \overline{f}(t)dt = 0$. Then $\overline{f}dt = d \overline{g}$ for a $C^{\infty}$ periodic function $\overline{g}$ of period $2\pi$ on $\mathbb{R}$.
I will not write the proof of the lemma. The example then continues
Let $\overline{g}$ be the periodic function of period $2\pi$ on $\mathbb{R}$ from lemma 24.5. By proposition 18.12 $\overline{g} = h^* g$ for some $C^{\infty}$ function $g$ on $S^1$. It follows that $$ d \overline{g} = dh^* g = h^*(dg) $$ Since $h^* :\Omega^1(S^1) \to \Omega^1(\mathbb{R})$ is injective, $\alpha = d g$.
Here I got a bit confused on the injectivity, probably trivial. I did this
$$ h^*(f_1 dx + g_1 dy) = h^*(f_2 dx + g_2 dy) \iff h^*f_1 dh^*x + h^*g_1 dh^*y = h^*f_2 dh^*x + h^*g_2 dh^*y \iff (-(f_1 \circ h)(t) \sin(t) + (g_1 \circ h)(t) \cos(t)) dt = (-(f_2 \circ h)(t) \sin(t) + (g_2 \circ h)(t) \cos(t)) dt \iff (-(f_2 \circ h)(t) + (f_1 \circ h)(t)) \sin(t) + (-(g_1 \circ h)(t) + (g_2 \circ h)(t)) \cos(t) = (f_1 - f_2) \circ h)(t) \sin(t) + (g_2 - g_1) \circ h)(t)\cos(t) = 0 $$
Last equation implies the following system needs to be satisfied for all $t$
$$ \left\{ \begin{array}{l} f_1 - f_2 = 0 \\ g_1 - g_2 = 0 \end{array} \right. $$
So $f_1 = f_2$ and $g_1 = g_2$ so we have injectivity. Is this correct?
This proves that the kernel of $\varphi$ consist of exact forms. Therefore, integration induces an isomorphism $$ H^1(S^1) = \frac{Z^1(S^1)}{B^1(S^1)} = \mathbb{R}. $$
Why is in this case the quotient space isomorphic to $\mathbb{R}$? Are we using the first isomorphism theorem maybe? I got confused?
If you are new to cohomotopy and you want to understand where these calculations come from, I'll give you some clues and I think you can Answer to your sub-questions:
Why calculation of the cohomology of circle is a little long and unusual and where these calculations come from?
because $Z^1(M)$ is 1-dimensional vector space (over ?) so by finding a non-zero differential form $\omega$ (as a base) you can produce all the space. So everything else is of the form $\alpha = \lambda \omega$.
Your comment is true for surjectivity. I think everything else is clear. isn't?