I am working on the following proof:
Let $V$ and $W$ be finite-dimensional vector spaces and let $f: V \longrightarrow W$ be linear. Proof that $Df(x) = f$.
Now, I have seen some proofs on the internet that use the fact, that for each linear function $f$ there exists a matrix $A$ such that forall $x \in V \, \exists \, y \in W$ so that
$$y = A \cdot x$$
or componentwise
$$y_i = A_{ij} \, x_j$$
Therefore, one has
$$ \frac{\partial y_i}{\partial x_k} = A_{ij} \, \delta_{jk} = A_{ik} $$
Which would prove the identity $Df(x) = f$. Now, my problem with this proof is, that the derivatives
$$\frac{\partial y_i}{\partial x_k}$$
should be understood as the derivatives of $y_i$ with respect to the coordinates $x_k$. The equation $y = A \cdot x$ however takes a vector $x \in V$ and maps it onto a vector $y \in W$. So, calling the vector $x$ is just a name for an object in $V$, which I could also have called $v \in V$. Therefore the derivatives
$$\frac{\partial x_i}{\partial x_k}$$
are actually
$$\frac{\partial v_i}{\partial x_k}$$
and therefore not (necessarily) $\delta_{ik}$. What do I miss here? Am I misconceptioning something here?
EDIT 1
With EXODD's help I figured out the proof as follows (see comments).
$$\partial_k f_i(v) = f_i(e_k) = \tilde{e}_i \cdot f(e_k)$$
where $e_i$ are a basis of $V$ and $\tilde{e}_i$ are a basis of $W$. Now, there exists a transformation matrix $A$ such that
$$\tilde{e}_i \cdot f(e_k) = A_{ik}$$
which in turn are the components of the transformation matrix for the linear map $f$. Thank you for your help EXODD.
${}$
However, I am again stuck with the follow-up-question. It is
Proposition: Let $U, V, W$ be finite-dimensional vector spaces and let $g : U \times V \longrightarrow W$ be bilinear. Show that for $(u_1,v_1) \in U \times V$ and $(u_2, v_2) \in U \times V$ it holds that
$$\Big( D \, g(u_1,v_1) \Big) (u_2, v_2) = g(u_1,v_2) + g(u_2,v_1)$$.
Proof: By calculating the partial derivative, I came up with
\begin{align*} \frac{\partial}{\partial x_j} f_i(u_1,v_1) &= \lim_{h \rightarrow 0} \frac{1}{h} \Big( f_i (u_1 + he_j, v_1 + he_j) - f_i(u_1, v_1) \Big) \\ &{} \\ &= \lim_{h \rightarrow 0} \frac{1}{h} \Big( hf_i(u_1, e_j) + hf_i(e_j,v_1) + h^2f_i(e_j,e_j) \Big) \\ &{} \\ &= f_i(u_1, e_j) + f_i(e_j,v_1) \end{align*}
But again, I am unable to proceed from here.