Coordinate independence of connections So I am trying to prove the following: 

Let $V \rightarrow M$ be a vector bundle $\nabla$ a connection on $V$. Then there is a unique sequence of linear maps 
      $$ \Omega^0(M;V) \xrightarrow{\nabla} \Omega^1(M;V) \xrightarrow{\nabla} \cdots $$ 
   such that $\nabla$ coincdies with the connection for $p=0$ and such that $$\nabla (w \wedge s ) = dw \wedge  s + (-1)^{|w|} w \wedge \nabla s  (*) $$

Where $\Omega^k(M;V) := \Omega^k(M) \otimes_{\Omega^0(M)} \Omega^0(V)$, where $\Omega^k(M)$ are the smooth $k$-forms ($k=0$ give smooth functions), and $\Omega^0(V)$ are the smooth sections of $V \rightarrow M$. 
A connection $\nabla:\Omega^0(M;V) \rightarrow \Omega^1(M;V)$ is defined to be a map that satisfies 
$$ \nabla (fs) = df \otimes s + f \nabla s $$ 


So I wanted to define $\nabla$ locally. Since given a local frame $e_i$ for $V$, we may write $s = \sum w_i \otimes e_i$ and  we must have
$$ \nabla s = \sum_i dw_i \otimes e_i  + w _i \wedge \nabla  e_i $$ 


The problem is I could not show this is coordinate independent. 

My failed attempt: 
given another local frame $f_1, \ldots, f_r$ of $V$. Suppose $e = Af$, so $e_i  = \sum A_{ij} f_j $ where $A_{ij} \in C^\infty(U)$. 
Then 
\begin{align*}
\sum_i \Big(dw_i \otimes \sum A_{ij} f_j +w_i \wedge \nabla (\sum A_{ij} f_j )  \Big)  &= \sum_i  \Big(\sum_j   A_{ij}  dw_i \otimes f_j+  w_i \wedge (\sum_j dA_{ij} \otimes f_j + A_{ij}  \nabla f_j  ) \Big) 
\end{align*} 

Since $ A_{ij} dw_i + w_i dA_{ij} = d(w_i A_{ij})$ for all $i$ and $j$, the expression on the right-hand side reduces to
$$ \sum_{i}\sum_j \left( d(w_i A_{ij}) \otimes f_j + (w_i A_{ij}) \nabla f_j \right),$$
which is precisely the expression you want.

Edit: If the $w_i$'s are $k$-forms (rather than just smooth functions), then you would need to introduce an extra sign in your proposed definition:
$$ \nabla s = \sum_i dw_i \otimes e_i + (-1)^k w_i \wedge \nabla e_i.$$
You'll end up with a corresponding factor of $(-1)^k$ on the right-hand side of your equation. Fortunately, everything works out because $A_{ij} dw_i + (-1)^k w_i \wedge dA_{ij} = d(w_iA_{ij})$.
