Let M be as smooth manifold, W a k-dimensional vector space with basis $(e_1,...,e_k)$ and $\Omega^r(M,W)$ be the differential r-form on M with values in W. Let $\omega \in \Omega^r$ and $v_1,..., v_r \in T_xM$. Then we can write $\omega_x(v_1,...,v_r)$ as a linear combination of the $e_i$ as follows: $w_x(v_1,...,v_r) = \omega_x^i(v_1,...,v_r)e_i$, i.e. both $\omega_x$ and $\omega_x^i$ are multilinear maps and $\omega_i$ is a differential r-form on M. Let's now define the exterior differential as a sheaf morphism from $\Omega^r(M,W) \rightarrow \Omega^{r+1}(M,W)$:
$$d(\omega^i\otimes w_i):=d\omega_i\otimes e_i.$$
My question is now why this is independent of the basis we have chosen for W?
The reason is that the change of basis is given by constants. If you take a basis $\{f_j\}$ then you get an invertible matrix $A=(a_j^i)$ charakterized by the fact that $f_j=\sum_i a_j^ie_i$. Then writing $\omega$ as $\sum_j\tilde\omega^j\otimes f_j$ we see that $\omega=\sum_i\omega^i\otimes e_i$, where $\omega^i=\sum_j a^j_i\tilde\omega^j$. But since the $a^j_i$ are constants, this implies $d\omega^i=\sum_j a^j_id\tilde\omega^j$ and so $\sum_id\omega^i\otimes e_i=\sum_{i,j}a^j_id\tilde\omega^j\otimes e_i$. But then the constants can be brought to the other factor in the tensor product (the right one) and the $d\tilde\omega^j$ can be taken out of the sum over $i$ to obtain $\sum_jd\tilde\omega^j\otimes(\sum_ia^i_je_i)=\sum_j d\tilde\omega^j\otimes f_j$.