I want to prove that:
$$\phi_\beta:\mathsf{V}\to\mathsf{F}^n,\phi_\beta(x)=[x]_\beta,$$
which $\phi_\beta$ is called the standard representation of n-dimensional vector space $\mathsf{V}$ with respect to an ordered basis $\beta$. And $x\in\mathsf{V}$.
My proof:
(1.) Show $\phi_\beta$ is linear:
$\phi_\beta(cx+y)=?$
(2.) Show it's one-to-one and onto:
Let $\dim(\mathsf{V})=n.$ Let $u,v\in\mathsf{V}$ and $[u]_\beta=[v]_\beta$. Since $\beta$ is a basis of $\mathsf{V}, u=v$. So $\phi_\beta$ is one-to-one.
Also, since $\beta$ is a basis of $\mathsf{V}$, for any given $x\in \mathsf{F}^n$, $[x]_\beta$ uniquely express a vector in $\mathsf{V}$.
I completely don't know how to continue (1.), the only evidence I have is $\beta$ is an ordered basis. And for (2.) I'm proving it like I already know it, the only thing I used is $\beta$ an ordered basis again.
I'll get you started with the proof that $\phi_\beta(x+y)=\phi_\beta(x)+\phi_\beta(y)$. Let $\beta=(b_1,\dots,b_n)$.
Suppose $\phi_\beta(x)=(c_1,c_2,\dots,c_n)$ and $\phi_\beta(y)=(d_1,d_2,\dots,d_n)$. By definition of $\phi_\beta$, this means $x$ and $y$ can be written as $x = c_1b_1+c_2b_2+\dots+c_nb_n$ and $y=d_1b_1+d_2b_2+\dots+d_nb_n$. Therefore, $$ x+y = (c_1+d_1)b_1+\dots+(c_n+d_n)b_n $$ which proves that $$\phi_\beta(x+y)=(c_1+d_1,c_2+d_2,\dots,c_n+d_n)=(c_1,\dots,c_n)+(d_1,\dots,d_n)=\phi_\beta(x)+\phi_\beta(y)$$
In your proof of one-to-one and onto, you use the fact that $\beta$ is a basis to conclude things, but this is actually not necessary.
Suppose $[u]_\beta=[v]_\beta$. Let $[u]_\beta=(c_1,\dots,c_n)=[v]_\beta$. This means that $u=c_1b_1+\dots+c_nb_n$, and $v=c_1b_1+\dots+c_nb_n$, so therefore $u=v$. Notice that you do not need the fact that $\beta$ is a basis!
For onto, you need to show that for each $(c_1,\dots,c_n)\in \mathsf F^n$, there exists a $u$ for which $\phi_\beta(u)=(c_1,\dots,c_n)$. But that is easy, just let $u = c_1b_1+\dots+c_nb_n$.
So where does $\beta$ being a basis come in? You need the fact that $\beta$ is a basis to prove that the map $\phi_\beta$ exists in the first place, because this is what allows you to write $v$ as a linear combination of elements of $\beta$ in a unique way.