Any $1$-form (only on $S^3$?) can be uniquely written as a sum of a closed $1$-form and a coclosed $1$-form? What is meant by saying that any $1$-form (only on $S^3$?) can be uniquely written as a sum of a closed $1$-form and a "co-closed" $1$-form?
[...Since $H^1$ of $S^3$ is trivial it follows that the required closed $1$-form is also exact and then for any $1$-form say $A$ (a gauge field) on $S^3$ one says that it uniquely determines a gauge fixed 1-form $B$ such that $A = d\phi + B+\cdots$ ]
 A: Let $M$ be a compact Riemannian manifold. The Hodge Theorem states that for every $k$
$$\Omega^k(M) = d\Omega^{k-1}(M) \oplus \delta\Omega^{k+1}(M) \oplus \mathcal{H}^k(M)\quad \text{and}\quad H^k_{\text{dR}}(M) \cong \mathcal{H}^k(M)$$ 
where $\delta$ is the adjoint of $d$ and $\mathcal{H}^k(M) = \ker \Delta|_{\Omega^k(M)}$ is the space of harmonic $k$-forms. Therefore, any $k$-form $\omega$ on $M$ can be written uniquely as $\omega = d\alpha + \delta\beta + \gamma$ where $\alpha \in \Omega^{k-1}(M)$, $\beta \in \Omega^{k+1}(M)$, and $\gamma \in \mathcal{H}^k(M)$. As $d^2 = 0$ and $\delta^2 = 0$, $d\alpha$ is closed and $\delta\beta$ is coclosed. As $\gamma$ is harmonic, it is closed and coclosed (this follows from the fact that $(\Delta\gamma, \gamma) = \|d\gamma\|^2 + \|\delta\gamma\|^2$). Using the fact that the sum of (co)closed forms is (co)closed, $\omega$ is a sum of a closed form and a coclosed form, but not necessarily in a unique way:
$$\omega = \underbrace{(d\alpha + \gamma)}_{\text{closed}} + \underbrace{\delta\beta}_{\text{coclosed}} = \underbrace{d\alpha}_{\text{closed}} + \underbrace{(\delta\beta + \gamma)}_{\text{coclosed}}.$$
So, if $\omega$ can be written uniquely as the sum of a closed form and a coclosed form, it is necessary that $\gamma = 0$. This condition is also sufficient by the Hodge Theorem. Therefore, every $k$-form on $M$ can be written uniquely as the sum of a closed form and a coclosed form if and only if $H^k_{\text{dR}}(M) \cong \mathcal{H}^k(M) = 0$.
Note: The above discussion follows through in the exact same way for vector bundle valued forms.

The statement you are referring to is the specific case $M = S^3$, $k = 1$ which applies because $H^1_{\text{dR}}(S^3) = 0$.
