Suppose that $M$ is simply connected, compact manifold. Then I want to show that every closed $1-$ form is exact on $M$.
This is same showing that the first de-Rham cohomology space of $M$ is $0$.
Question: Is it possible to answer this without using universal coefficients theorem?
(The reason for not using the universal coefficients theorem is that it contains a mystery term ext which I'm not comfortable using. Digging deeper into ext, it turns out that it is a core algebraic term which I'm not good enough at at this stage.)
Since $M$ is simply connected, its fundamental group $\pi_1$ is $0$. First homology group $H_1$ is abeliazation of $\pi_1$ so $H_1 =0$. To conclude $H^1=0$, I would need universal coefficients theorem, which is what I want to avoid.
Does the following work?
Suppose that $\omega$ is a closed $1$ form on $M$. Consider any $p$ in M. Take a chart at $p$ and call it $h$. Then $\tau=(h^{-1})^\ast \omega$ is a $1$ form on $\mathbb R^n$ hence equal to $df$ for some smooth $f$ on $\mathbb R^n$. Pulling $\tau$ back to $M$ and using that fact that $d$ and $\ast$ commute, I have $\omega= d(h^\ast f)$. This is true for every point $p$. Locally this is fine.
But here $h$ is not the same function throughout. $h$ depends upon $p$. So how to create a $0$ form $g$ on $M$ such that $\omega =dg$?
To answer your Question, note that one does not need universal coefficients to state at least some link between $\newcommand{\HdR}{H_\mathrm{dR}}\HdR^1(M)$ and $\pi_1(M)$. Namely, you can define a map $\Phi\colon \HdR^1(M) \to \operatorname{Hom}_{\mathrm{Grp}}(\pi_1(M, x), \mathbb{R})$ (where $x \in M$ is any point) via $$ \Phi([\omega])([\gamma]) = \int_{\tilde{\gamma}} \omega $$ where $\tilde{\gamma} \in [\gamma]$ is a smooth representative (here the integral is the usual line integral, i.e. it is given by $\int_{\tilde{\gamma}} \omega = \int_{[a, b]} \tilde{\gamma}^* \omega$ for $\tilde{\gamma}\colon [a, b] \to M$ a smooth curve). It turns out that this map is an isomorphism, and you can at least see directly that it is injective without resorting to de Rham's theorem and singular cohomology at all: In fact, if $\int_{\tilde{\gamma}} \omega = 0$ for all choices of $\tilde{\gamma}$, then it is not hard to see that $\int_\sigma \omega = 0$ for $\sigma$ a smooth closed curve based at another point $y \in M$, whence $\omega$ is conservative and therefore $[\omega] = 0$ (for some more details, see Lee's Introduction to Smooth Manifolds, Theorem 17.17). Obviously, then, if $\pi_1(M, x) = 0$, so, too, must be $\HdR^1(M)$.
...but of course, you don't really need to invoke the universal coefficient theorem to go from $H_1(M; \mathbb{Z}) = 0$ to $H^1(M; \mathbb{R}) \cong \HdR^1(M) = 0$ either! Namely, let $(C_\bullet(M), d_\bullet) = (\cdots \overset{d_2}{\to} C_1(M) \overset{d_1}{\to} C_0(M))$ be the singular chain complex of $M$ computing $H_\bullet(M; \mathbb{Z})$. Note that $H_1(M; \mathbb{Z}) = 0$ is the same as saying that this complex is exact at $C_1(M)$, i.e. that $\operatorname{img} d_2 = \ker d_1$. To go from this to the cochain complex defining $H^\bullet(M; \mathbb{R})$, you by definition apply $\operatorname{Hom}({{-}}, \mathbb{R})$ to get $$ \operatorname{Hom}(C_0(M), \mathbb{R}) \overset{d_1^*}{\longrightarrow} \operatorname{Hom}(C_1(M), \mathbb{R}) \overset{d_2^*}{\longrightarrow} \cdots $$ and take its cohomology, but since $\mathbb{R}$ is injective, $\operatorname{Hom}({{-}}, \mathbb{R})$ is exact, which implies that $\operatorname{img} d_1^* = \ker d_2^*$, whence $H^1(M; \mathbb{R}) = 0$.
As for the question about explicitly finding a 0-form whose derivative is any given 1-form, I'll gladly leave it to somebody else :)