There are already quite a number of questions on basis change in a vector space. Nevertheless, to fully grasp the underlying idea I made up the following notation and I have some doubts on it (note: the questions are at the end...)
Let $K$ be a field and $V$ a $K$-vector space of dimension $n$. Let $\mathbf{v}=(v_1,v_2,...,v_n)\in V^n$ and $A\in M_{n\times m}(K)$ be a $K$-valued matrix (Note that if $V=\mathbb{R}^n$, $\mathbf{v}$ is a real-valued square matrix). We can define a product of $\mathbf{v}$ and $A$ in the following manner: \begin{equation} \star: V^n\times M_{n\times m}(K)\longrightarrow V^m\\ (\mathbf{v},A)\longmapsto \mathbf{v}\star A=\mathbf{w} \end{equation} with $\mathbf{w}=(w_1,...w_m)$ and $w_i=\sum_jv_jA_{ji}$ (remember, the $v_j$'s are vectors, not components- we don't need a basis for $V$ to compute this).
The idea behind all this is that with this notation the effect of a change of basis in $V$ on the components of a vector is quickly (and I think neatly) determined as follows.
Given a basis $\mathbf{v}\in V^n$ of $V$ and a vector $v\in V$, we can consider the map $\pi_{\mathbf{v}}:V\rightarrow K^n$ which gives the components of $v$ with respect to the basis $\mathbf{v}$, i.e. if $v=a_1v_1+...+a_nv_n$, then $\pi_{\mathbf{v}}(v)=(a_1,...,a_n)$. Having defined $\star$, for every vector $u\in V$ we can write \begin{equation} u=\mathbf{v}\star\pi_{\mathbf{v}}(u) \end{equation} If $\mathbf{w}$ is another basis for $V$, there is a square invertible matrix $A$ such that $\mathbf{v}\star A=\mathbf{w}$. Let's call $A$ the "change of basis matrix". Since we have $u=\mathbf{v}\star\pi_{\mathbf{v}}(u)$ and also $u=\mathbf{w}\star\pi_{\mathbf{w}}(u)$, it follows \begin{equation} u=\mathbf{w}\star\pi_{\mathbf{w}}(u)=(\mathbf{v}\star A)\star\pi_{\mathbf{w}}(u)= \mathbf{v}\star (A\cdot\pi_{\mathbf{w}}(u))=\mathbf{v}\star\pi_{\mathbf{v}}(u) \end{equation} where $A\cdot\pi_{\mathbf{w}}(u)$ is the usual matrix-vector product. Since $\mathbf{v}$ is a basis, from the last equality we deduce that \begin{equation} \pi_{\mathbf{w}}(u)=A^{-1}\cdot\pi_{\mathbf{v}}(u) \end{equation} which is a relation on the components of the vector $u$, in particular it shows that they transform in a contravariant fashion with respect to the tranformation of the basis.
This reasoning applies also to a change of basis in the dual $V^*$ of $V$, and the result I get is that if $\mathbf{v}^*$ and $\mathbf{w}^*$ are two bases for $V^*$ (dual to $\mathbf{v}$ and $\mathbf{w}$ repectively), $\pi_{\mathbf{v}^*}$ and $\pi_{\mathbf{w}^*}$ are the coordinate maps $V^*\rightarrow K^n$ defined as before and $u^*\in V^*$, then \begin{equation} \mathbf{w}=\mathbf{v}\star A\qquad\Longrightarrow\qquad \pi_{\mathbf{w}^*}(u^*)=A^T\cdot\pi_{\mathbf{v}^*}(u^*) \end{equation} ($A^T$ is the transpose of $A$). This should show that the components of a linear form transform in covariant fashion with respect to the basis transformation.
So, that was the idea. Finally the questions (actually question 2 is more important than question 1):
1) Are these calculations correct? (terrible question!)
2) Could this notation be extendable to bilinear forms and eventually multilinear forms? I am desperately trying to wrap my head around tensors but I need a deep understanding of what is going on in the "lower levels" first...