Theorem: $[T]_E^F[v]_E=[T(v)]_F$, where $T:F^n→F^m$ is a linear map I have written a proof for the theorem below. I would appreciate comments on where I have made mistakes. Also if you have any simpler proofs I would be grateful. Thanks in advance.
Theorem: $[T]_E^F[v]_E=[T(v)]_F$, where $T:F^n→F^m$ is a linear map

Proof. Part 1 $[T(v)]_F$
Step 1. $T(v)=Av$ by the theorem that any linear transformation can be represented by a matrix for $T:F^n→F^m$.
2. Take $E$={$e_1,\cdots, e_n$} as the standard basis of $F^n$, then for $v∈F^n$, $v=\alpha_1e_1+\cdots+\alpha_ne_n$=$$\begin{pmatrix}\alpha_1\\\vdots\\\alpha_n\\\end{pmatrix}$$
3.$$Av=\begin{pmatrix}a_{11}&\cdots&a_{1n}\\\vdots&\ddots&\vdots\\a_{m1}&\cdots&a_{mn}\\\end{pmatrix}\begin{pmatrix}\alpha_1\\\vdots\\\alpha_n\\\end{pmatrix}=\begin{pmatrix}\alpha_1a_{11}+\cdots+\alpha_na_{1n}\\\vdots\\\alpha_1a_{m1}+\cdots+\alpha_na_{mn}\\\end{pmatrix}$$
4. Take $F$={$f_1,\cdots, f_m$} as the standard basis of $F^m$ where $w∈F^m$, then $w=[w]_F$. As $w=Av$, $Av=[Av]_F$,  therefore:$$[Av]_F=\begin{pmatrix}\alpha_1a_{11}+\cdots+\alpha_na_{1n}\\\vdots\\\alpha_1a_{m1}+\cdots+\alpha_na_{mn}\\\end{pmatrix}$$
5. As $Av=T(v)$: $$[T(v)]_F=\begin{pmatrix}\alpha_1a_{11}+\cdots+\alpha_na_{1n}\\\vdots\\\alpha_1a_{m1}+\cdots+\alpha_na_{mn}\\\end{pmatrix}$$
Part 2 $[T]_E^F[v]_E$
Step 1. As stated, $E$ is the standard basis of $F^n$ and $F$ is the standard basis of $F^m$. $T$ is a linear map, $T:F^n→F^m$
2. If we apply $T$ to the basis vectors of $F^n$ then we get an element in $F^m$ which is a linear combination of the basis vectors of $F^m$. Therefore:
$$T(e_1)=a_{11}f_1+a_{21}f_2+\cdots+a_{m1}f_m\\\vdots\\T(e_n)=a_{1n}f_1+a_{2n}f_2+\cdots+a_{mn}f_m$$
3. The above system can be represented in matrix form:
$$T\begin{pmatrix}e_1\\\vdots\\e_n\end{pmatrix}=\begin{pmatrix}a_{11}f_1+a_{21}f_2+\cdots+a_{m1}f_m\\\vdots\\a_{1n}f_1+a_{2n}f_2+\cdots+a_{mn}f_m\end{pmatrix}$$
$$=\begin{pmatrix}a_{11}f_1+a_{21}f_2+\cdots+a_{m1}f_m&\cdots&a_{1n}f_1+a_{2n}f_2+\cdots+a_{mn}f_m\end{pmatrix}$$
$$=\begin{pmatrix}f_1&\cdots&f_m\end{pmatrix}\begin{pmatrix}a_{11}&\cdots&a_{1n}\\\vdots&\ddots&\vdots\\a_{m1}&\cdots&a_{mn}\\\end{pmatrix}$$
4. Therefore: $$[A]_F=\begin{pmatrix}a_{11}&\cdots&a_{1n}\\\vdots&\ddots&\vdots\\a_{m1}&\cdots&a_{mn}\\\end{pmatrix}$$
5. As both E and F are standard bases, $[A]_F=A=[A]^F_E$ and any linear transformation can be represented by a matrix, therefore $[A]^F_E=[T]^F_E$
6. Therefore:$$[T]^F_E[v]_E=[A]^F_E[v]_E=A[v]_E=\begin{pmatrix}a_{11}&\cdots&a_{1n}\\\vdots&\ddots&\vdots\\a_{m1}&\cdots&a_{mn}\\\end{pmatrix}\begin{pmatrix}\alpha_1\\\vdots\\\alpha_n\\\end{pmatrix}$$$$=\begin{pmatrix}\alpha_1a_{11}+\cdots+\alpha_na_{1n}\\\vdots\\\alpha_1a_{m1}+\cdots+\alpha_na_{mn}\\\end{pmatrix}$$
Therefore $[T]^F_E[v]_E=[T(v)]_F$
Q.E.D.
 A: Your very first step is wrong. You can't represent any linear map $\textsf T : \textsf V \to \textsf W$ by $\textsf T (v) = Av$, since we don't even know if $v$ is a column vector to be able to perform the matrix multiplication. 
In fact, if you already know that $[ \textsf U \circ \textsf T ]_\alpha^\gamma = [ \textsf U ]_\beta^\gamma [ \textsf T ]_\alpha^\beta$ for any linear transformations $\textsf T$ and $\textsf U$, whenever that its composition its possible, it is much easier to think about it as follows.

Theorem. Let $\textsf V$ and $\textsf W$ be finite-dimensional vector spaces over a field $F$, having ordered basis $\beta$ and $\gamma$, respectively. Let $\textsf T : \textsf V \to \textsf W$ be linear. Then for each $u \in \textsf V$ we have $$[\textsf T (u)]_\gamma = [\textsf T]_\beta^\gamma [u]_\beta$$

Proof : Let $u \in \textsf V$. Define $f : F \to \textsf V$ and $g : F \to \textsf W$ by $f(a)=au$ and $g(a)=a\textsf T(u)$ for all $a\in F$. 
Clearly, $f$ and $g$ are linear transformations and we have that $g=\textsf T \circ f$. Let $\alpha = \{1\}$ be the standard ordered basis for $F$, and from what I mentioned above, we get that
$$\begin{align}
[ \textsf T(u) ]_\gamma &= [g(1)]_\gamma = [g]_\alpha^\gamma \\
&= [ \textsf T \circ f ]_\alpha^\gamma = [ \textsf T]_\beta^\gamma [f]_\alpha^\beta \\
&= [ \textsf T]_\beta^\gamma [f(1)]_\beta = [\textsf T]_\beta^\gamma [u]_\beta
\end{align}$$
