Let $\pi:P\to M$ be a principal $G$-bundle with some connection $\omega\in\Omega^1(P,\mathfrak g)$. Let $\psi\in \Omega^k(P)$ such that $\psi_{p\cdot g}\big(r_{g*}(v_1),...,r_{g*}(v_k)\big)=\psi_p(v_1,...,v_k)$ for all $g\in G$ and all $p\in P$ with any $v_1,...,v_k\in T_pP$ i.e. $\psi$ is $G$-invariant. Here, $r_g:P\to P$ is defined by $p\mapsto p\cdot g$. Further, suppose, $\psi$ is vanishes whenever one of its argument is a vertical vector. I want to show, $\psi$ is a basic form i.e. $\psi\in \text{Im}\big(\pi^*:\Omega^k( M)\to \Omega^k(P)\big)$.
So, define $\psi^\flat\in \Omega^k( M)$ as follows: choose $m\in M$ and $u_1,...,u_k\in T_m M$, then define $\psi^\flat_m(u_1,...,u_k):=\psi_p(v_1,...,v_k)\in \Bbb R$, where $\pi(p)=m$ and $\pi_{*p}(v_i)=u_i$ for all $i$. Now we show $\psi^\flat$ is well-defined. Choose another set $v_1',...,v_k'$ with $\pi_{*p}(v_i')=u_i$ for all $i$, then $v_i-v_i'$ are vertical vectors for all $i$. Hence, $\psi_p(v_1',...,v_k')=\psi_p\big(v_1+\text{vertical},...,v_k+\text{vertical}\big)=\psi_p(v_1,...,v_k)$. Next, choose $q\in \pi^{-1}(m)$, so that $q=p\cdot g$ for some $g\in G$. Then, $\pi_{*q}\big(r_{g*}(v_i)\big)=\big(\pi\circ r_g)_{*p}(v_i)=\pi_{*p}(v_i)=u_i$ for all $i$, hence using $G$-invariance of $\psi$ we have$\psi_q\big(r_{g*}(v_1),...,r_{g*}(v_k)\big)=\psi_p(v_1,...,v_k)$. So we are done. Also, $\pi^*\big(\psi^\flat\big)=\psi$, i.e. $\psi^\flat_m\big(\pi_{*p}(v_1),...,\pi_{*p}(v_k)\big)=\psi_p(v_1,...,v_k)$. So $\psi^\flat$ is well defined.
Now, my question is how to show $\psi^\flat$ is smooth, that is given any vector fields $Z_1,...,Z_k\in \mathfrak X(M)$ we have to show $M\ni m\mapsto \psi^\flat_m\big(Z_1(m),...,Z_k(m)\big)\in\Bbb R$ is smooth. I already know, horizontal lifting lemma, i.e. given $Z_1,...,Z_k\in\mathfrak X(M)$ we have horizontal vector fields $\widetilde Z_1,...,\widetilde Z_k\in\mathfrak X(P)$ such that $\pi_{*p}\big(\widetilde Z_i(p)\big)=Z_i\big(\pi(p)\big)$ for all $p\in P$. I guess that I have to use the fact that $\pi$ is a surjective submersion. Any help will be appricted.
I think you've already done the hard part and you do not need the horizontal lifting lemma to check smoothness. First, to check that $\psi^\flat$ is smooth, it suffices to check on an open cover of $M$, and we can pick a cover trivializing the bundle.
Then on an open set $U$, we can pick a smooth section $s_U$ of the projection $\pi$, that is, a map with $\pi \circ s_U = \text{id}_U$ (we are using that $U$ is small here as there may be no global sections). Now on $U$ we have $$ \psi^\flat = \psi \circ {s_U}_* $$ and so $\psi^\flat$ is a composition of smooth maps.