[proofplan]
We compare two principal connection forms by subtracting them as $\mathfrak g$-valued $1$-forms on $P$. The reproduction axiom on fundamental vertical vector fields forces the difference to be horizontal, and the equivariance axiom forces it to transform by the adjoint representation. These two properties are precisely the standard identification of $\Omega^1(M;\operatorname{ad}(P))$ with horizontal, $G$-equivariant $\mathfrak g$-valued $1$-forms on $P$. Conversely, adding such a horizontal equivariant form to a fixed connection preserves exactly the two defining axioms of a principal connection.
[/proofplan]
[step:Identify adjoint-valued forms with horizontal equivariant forms on $P$]
Let $\Omega^1_{\mathrm{hor}}(P;\mathfrak g)^G$ denote the [vector space](/page/Vector%20Space) of smooth $\mathfrak g$-valued $1$-forms $\beta \in \Omega^1(P;\mathfrak g)$ satisfying the following two conditions. First, $\beta$ is horizontal:
\begin{align*}
\beta_p(v) = 0
\end{align*}
for every $p \in P$ and every vertical tangent vector $v \in \ker d\pi_p$. Second, $\beta$ is $G$-equivariant:
\begin{align*}
\beta_{pg}(dR_g|_p(v)) = \operatorname{Ad}_{g^{-1}}(\beta_p(v))
\end{align*}
for every $p \in P$, $g \in G$, and $v \in T_pP$.
We use the standard interpretation
\begin{align*}
\Omega^1(M;\operatorname{ad}(P)) \cong \Omega^1_{\mathrm{hor}}(P;\mathfrak g)^G
\end{align*}
where $\operatorname{ad}(P)=P\times_G\mathfrak g$ is the adjoint bundle associated to the adjoint action convention encoded by the above equivariance law. Under this identification, an adjoint-valued $1$-form $\alpha \in \Omega^1(M;\operatorname{ad}(P))$ is represented by a unique form $\widetilde{\alpha} \in \Omega^1_{\mathrm{hor}}(P;\mathfrak g)^G$.
[guided]
We first isolate the linear object that will model the affine space. Define $\Omega^1_{\mathrm{hor}}(P;\mathfrak g)^G$ to be the vector space of smooth $\mathfrak g$-valued $1$-forms $\beta: TP \to \mathfrak g$ with two properties.
The first property is horizontality. For every $p \in P$ and every vertical vector $v \in \ker d\pi_p$, the value of $\beta$ must vanish:
\begin{align*}
\beta_p(v) = 0.
\end{align*}
This condition says that $\beta$ only records directions tangent to the base manifold $M$.
The second property is equivariance under the right principal action. If $R_g: P \to P$ is the right action by $g \in G$, then for every $p \in P$ and $v \in T_pP$ we require
\begin{align*}
\beta_{pg}(dR_g|_p(v)) = \operatorname{Ad}_{g^{-1}}(\beta_p(v)).
\end{align*}
This condition is exactly what makes the value of $\beta$ independent of the choice of frame $p$ over a base point, once the value is interpreted in the associated adjoint bundle.
With this convention, the associated adjoint bundle is $\operatorname{ad}(P)=P\times_G\mathfrak g$, and the standard correspondence identifies adjoint-valued $1$-forms on $M$ with horizontal, $G$-equivariant $\mathfrak g$-valued $1$-forms on $P$:
\begin{align*}
\Omega^1(M;\operatorname{ad}(P)) \cong \Omega^1_{\mathrm{hor}}(P;\mathfrak g)^G.
\end{align*}
Thus, whenever $\alpha \in \Omega^1(M;\operatorname{ad}(P))$ appears in the formula $\omega_0+\alpha$, the actual form being added on $P$ is its unique representative $\widetilde{\alpha} \in \Omega^1_{\mathrm{hor}}(P;\mathfrak g)^G$.
[/guided]
[/step]
[step:Show that the difference of two connections is horizontal and equivariant]
Let $\omega_0,\omega_1 \in \mathcal A(P)$, and define the smooth $\mathfrak g$-valued $1$-form
\begin{align*}
\beta: TP \to \mathfrak g,\quad \beta_p(v)=\omega_{1,p}(v)-\omega_{0,p}(v).
\end{align*}
For $\xi \in \mathfrak g$, let $\xi_P: P \to TP$ be the fundamental vertical vector field generated by $\xi$. Since both $\omega_0$ and $\omega_1$ reproduce fundamental vertical vectors,
\begin{align*}
\beta_p(\xi_P(p)) = \omega_{1,p}(\xi_P(p))-\omega_{0,p}(\xi_P(p)) = \xi-\xi = 0.
\end{align*}
Every vertical tangent vector in $\ker d\pi_p$ is of the form $\xi_P(p)$ for a unique $\xi \in \mathfrak g$, because $P$ is a principal $G$-bundle and the infinitesimal action identifies $\mathfrak g$ with the vertical tangent space at $p$. Hence $\beta$ is horizontal.
For $g \in G$, the equivariance axiom for $\omega_0$ and $\omega_1$ gives
\begin{align*}
(R_g)^*\beta = (R_g)^*\omega_1-(R_g)^*\omega_0 = \operatorname{Ad}_{g^{-1}}\omega_1-\operatorname{Ad}_{g^{-1}}\omega_0 = \operatorname{Ad}_{g^{-1}}\beta.
\end{align*}
Therefore $\beta \in \Omega^1_{\mathrm{hor}}(P;\mathfrak g)^G$. By the identification from the previous step, there is a unique $\alpha \in \Omega^1(M;\operatorname{ad}(P))$ with $\widetilde{\alpha}=\beta$, and therefore
\begin{align*}
\omega_1 = \omega_0+\widetilde{\alpha}.
\end{align*}
[/step]
[step:Show that adding an adjoint-valued form gives another connection]
Let $\alpha \in \Omega^1(M;\operatorname{ad}(P))$, and let $\widetilde{\alpha} \in \Omega^1_{\mathrm{hor}}(P;\mathfrak g)^G$ be its horizontal equivariant representative. Define the smooth $\mathfrak g$-valued $1$-form
\begin{align*}
\omega_\alpha: TP \to \mathfrak g,\quad (\omega_\alpha)_p(v)=\omega_{0,p}(v)+\widetilde{\alpha}_p(v).
\end{align*}
For $\xi \in \mathfrak g$, the vector $\xi_P(p)$ is vertical, so horizontality of $\widetilde{\alpha}$ gives
\begin{align*}
(\omega_\alpha)_p(\xi_P(p)) = \omega_{0,p}(\xi_P(p))+\widetilde{\alpha}_p(\xi_P(p)) = \xi+0 = \xi.
\end{align*}
Thus $\omega_\alpha$ satisfies the reproduction axiom.
For $g \in G$, equivariance of $\omega_0$ and $\widetilde{\alpha}$ gives
\begin{align*}
(R_g)^*\omega_\alpha = (R_g)^*\omega_0+(R_g)^*\widetilde{\alpha} = \operatorname{Ad}_{g^{-1}}\omega_0+\operatorname{Ad}_{g^{-1}}\widetilde{\alpha} = \operatorname{Ad}_{g^{-1}}\omega_\alpha.
\end{align*}
Thus $\omega_\alpha$ satisfies the principal equivariance axiom. Hence $\omega_\alpha=\omega_0+\widetilde{\alpha}$ is a smooth principal connection form on $P$.
[/step]
[step:Verify freeness and transitivity of the affine action]
Define the translation map
\begin{align*}
T_{\omega_0}: \Omega^1(M;\operatorname{ad}(P)) \to \mathcal A(P),\quad T_{\omega_0}(\alpha)=\omega_0+\widetilde{\alpha}.
\end{align*}
The previous step shows that $T_{\omega_0}$ is well-defined. The difference step shows that every $\omega_1 \in \mathcal A(P)$ lies in the image of $T_{\omega_0}$, so $T_{\omega_0}$ is surjective.
If $T_{\omega_0}(\alpha)=T_{\omega_0}(\gamma)$ for $\alpha,\gamma \in \Omega^1(M;\operatorname{ad}(P))$, then
\begin{align*}
\omega_0+\widetilde{\alpha}=\omega_0+\widetilde{\gamma}.
\end{align*}
Subtracting $\omega_0$ as a $\mathfrak g$-valued $1$-form on $P$ gives $\widetilde{\alpha}=\widetilde{\gamma}$. By uniqueness in the identification of adjoint-valued forms with horizontal equivariant forms, $\alpha=\gamma$. Hence $T_{\omega_0}$ is injective.
Therefore $T_{\omega_0}$ is a bijection, and the additive vector space $\Omega^1(M;\operatorname{ad}(P))$ acts freely and transitively on $\mathcal A(P)$ by
\begin{align*}
\alpha \cdot \omega = \omega+\widetilde{\alpha}.
\end{align*}
This is precisely the statement that $\mathcal A(P)$ is an affine space modelled on $\Omega^1(M;\operatorname{ad}(P))$.
[/step]
