Show that a $k$-form $\omega$ is smooth if only if it is smooth as map $\omega : M\rightarrow \Lambda ^k(M)$ Let $M$ be a smooth manifold. Consider $\Lambda ^k(M)=\bigcup_{p\in M}\Lambda ^k(T_{p}M)$ with the natural smooth structure. With this structure I showed that the $\pi :\Lambda ^k(M)\rightarrow M$ projection is smooth.
Show that a $k$-form $\omega$ is smooth if only if it is smooth as map $\omega : M\rightarrow \Lambda ^k(M)$.
 A: $\Lambda^kM$ is a smooth manifold in its own right and hence you can ask
for a map $f\colon M \rightarrow \Lambda^kM$ to be smooth. Further such a map is called section of $\pi$, if $\pi (f(x))=x$ for all $x\in M$. 
Now a $k$-form $\omega$ can be seen as the map $$M\rightarrow \Lambda^kM, \quad x\mapsto \omega_x \in \Lambda_x^kM,$$
This map is obviuosly a section and your task is to prove that this section (read: "this map, which happens to be a section") is smooth if and only if $\omega$ is a smooth differential form. (What you actually have to do depends on how you have defined what a smooth differential form is. One common definition is that $\omega(X_1,\dots,X_k)$ is smooth for every collection $X_1,\dots,X_k$ of smooth vector fields)

EDIT:
Assume $M$ has dimension $n$, then $(d \phi_p^*)^{-1}\colon \Lambda_p^kM \rightarrow \mathbb{R}^{n \choose k}$. A chart of $\Lambda^kM$ rather looks like $(\phi,(d \phi^*)^{-1})\colon U \rightarrow \mathbb{R}^n \times \mathbb{R}^{n \choose k}$. Let $$F:= (\phi,(d \phi^*)^{-1}) \circ \omega \circ \phi^{-1}\colon \mathbb{R}^n \rightarrow  \mathbb{R}^n \times \mathbb{R}^{n \choose k},$$
then $F(x)=(x,(d \phi_p^*)^{-1}[\omega_p])$, where $\phi(p)=x$. Further
$$
(d\phi_p^*)^{-1}=d(\phi^{-1})_x^*,
$$
so in order for $F$ to be smooth you need that
$$ \mathbb{R}^n \ni 
x \mapsto d(\phi^{-1})_x^*\omega_{\phi^{-1}(x)}=\sum_K a_K(x) dx^K \in \mathbb{R}^{n\choose k}
$$
is smooth. ($K$ runs through ordered multi indices and if $K=(i_1,\dots,i_k)$, then $dx^K=dx_1^{i_1}\wedge \dots \wedge dx_1^{i_k}$). For this it is enough that all $a_K$ are smooth. Test this by plugging in vector fields $\frac{\partial}{\partial x_j}$.
