A simple way to think about this is to identify $V$ and $V^*$. Then the proof is slick, but it conceals the difference between $V$ and $V^*$, and the dependence of their identification on the dot product:
$$(S^Tx,x)=(x,Sx)=(Sx,x)>0.$$
In some cases, e.g. in tensor calculus and differential geometry, it is sometimes important to make this identification explicit (it corresponds to raising and lowering indices in coordinate notation). To make it explicit, it is convenient to think of $S^*:V^*\to V^*$ instead of $S^T$, and write $\langle\phi,x\rangle$ instead of $\phi(x)$ for the value of $\phi\in V^*$ on $x\in V$. The $\langle\cdot,\cdot\rangle$ is called the canonical pairing, and it saves us from writing a lot of nested parentheses. Then $S^*$ is defined by $\langle S^*\phi,x\rangle:=\langle\phi,S x\rangle$, and (a unique) functional $\phi$ is corresponded to $x$ via $\langle\phi,y\rangle=(x,y)$. There is also the induced inner product on $V^*$ with similar definition and properties. Now the calculation goes like this:
$$(S^*\phi,\phi)=\langle S^*\phi,x\rangle=\langle\phi,S x\rangle=(x,Sx)=(Sx,x)>0$$
There is a way to make this independent of an inner product. We should consider not operators on the same space, but operators from a space to its dual $S:V\to V^*$. They are used in tensor calculus (to contract indices), and in functional analysis, e.g. by Lions et al.
It is natural to call such operators positive definite when $\langle Sx,x\rangle>0$ for $x\neq0$. Defining $S^*$ by $\langle S^*x,y\rangle=\langle Sy,x\rangle$ we get $S^*:V^{**}\to V^*$. Note how neither inner product, nor basis, nor any other extra structure is needed so far. Moreover, $V$ can be identified with a subspace of its double dual $V^{**}$ also canonically, see injection into the double-dual. Namely, $x\mapsto\langle\cdot,x\rangle$ defines a functional on $V^*$ canonically corresponded to $x$. For finite-dimensional spaces $V$ maps onto the entire $V^{**}$. In general, when this happens the space is called reflexive. Now we recover the simplicity:
$$\langle S^*x,x\rangle=\langle Sx,x\rangle>0.$$
Matrices are useful representations of linear maps from one vector space to another (or the same one). But the transformation $\Psi : \mathbb{R}^2 \to \mathbb{R}^2$ from rectangular to polar coordinates, given by
$$ \newcommand{cif}{\mathrm{if}\ } \newcommand{cand}{\ \mathrm{and}\ } $$
$$ \Psi(x,y) = \left(\sqrt{x^2+y^2}, \Theta(x,y)\right) $$
$$ \Theta(x,y) = \begin{cases}
0 & \cif x=0 \cand y=0 \\
\arctan \frac{y}{x} & \cif x>0 \cand y>0 \\
\frac{\pi}{2} & \cif x=0 \cand y>0 \\
\pi + \arctan \frac{y}{x} & \cif x<0 \\
\frac{3 \pi}{2} & \cif x=0 \cand y<0 \\
2\pi + \arctan \frac{y}{x} & \cif x>0 \cand y<0 \\
\end{cases} $$
is not represented by a matrix because it is a non-linear transformation. Also, polar coordinates aren't a vector space: $(c r, c \theta)$ does not have a simple relationship to $(r,\theta)$, and $(r_1+r_2, \theta_1+\theta_2)$ does not have a simple relationship to $(r_1, \theta_1)$ and $(r_2, \theta_2)$. The image of $\Psi$ isn't even all of $\mathbb{R}^2$.
So most things wanting vector properties will just need to go back to the original coordinates, and in general there's no guarantee there will be a "nice" way to write them in a new coordinate system.
If $T$ is a linear transformation on $\mathbb{R}^2$ (rectangular), then its action on polar coordinates is:
$$ [T]_\Psi = \Psi \circ T \circ \Psi^{-1} $$
And we know the inverse $\Psi^{-1}$:
$$ \Psi^{-1}(r,\theta) = (r \cos \theta, r \sin \theta) $$
If we write
$$ T = \left(\begin{array}{cc} a & b \\ c & d \end{array}\right) $$
we get
$$ T\Psi^{-1}(r,\theta) = (ar\cos\theta + br\sin\theta, cr\cos\theta + dr\sin\theta) $$
So $\Psi T \Psi^{-1}(r,\theta)$ can be simplified a little bit, but isn't particularly pretty.
In a vector space, a norm acts as a distance function between vectors, $d(u,v) = \|u-v\|$, and obeys the triangle inequality $\|u+v\| \leq \|u\| + \|v\|$. This doesn't make as much sense in polar coordinates, where even adding or subtracting two points isn't exactly defined in the first place unless by going back to rectangular. But of course, if we just want to find the rectangular norm of a polar point, that's easy: $\|\Psi^{-1}(r,\theta)\| = r$. For other non-linear maps, it might not be so simple.
Polar coordinates again don't really have an inner product, since its properties related to multiplying by scalars and adding don't make direct sense. But the original rectangular inner product can be found as
$$ \begin{align*}
\left< \Psi^{-1}(r_1,\theta_1), \Psi^{-1}(r_2,\theta_2) \right> &=
\left< (r_1 \cos \theta_1, r_1 \sin \theta_1), (r_2 \cos \theta_2, r_2 \sin \theta_2) \right> \\
&= r_1 r_2 (\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2) \\
&= r_1 r_2 \cos(\theta_1 - \theta_2)
\end{align*} $$
which makes sense from knowing the dot product of two vectors in $\mathbb{R}^n$ is the product of their norms times the cosine of the angle between them.
There is one useful related generalization. For a function $f : \mathbb{R}^m \to \mathbb{R}^n$, we can define the derivative (or Jacobian matrix) $D_f$ as a function onto matrices, $D_f : \mathbb{R}^m \to \mathbb{R}_{n \times m}$, where the elements are the partial derivatives $\partial f_j(x_1,\ldots x_m)/\partial x_i$, because it has the property
$$ \forall x \in \mathbb{R^m}, u \in \mathbb{R^n}, v \in \mathbb{R^m} : \lim_{h \to 0} \frac{\langle u, f(x+hv) - f(x) \rangle}{h}\ = \langle u, D_f(x) v \rangle $$
It also follows a chain rule
$$ D_{f \circ g}(x) = D_f(g(x)) D_g(x) $$
For a linear map $T$, the derivative $D_T$ is a constant function whose value everywhere is the same matrix which ordinarily represents $T$. But in general, $D_f$ is different matrices at different points.
This derivative $D_f$ is important in seeing the effects of a change of coordinates on a multiple integral: Given a domain $S \subseteq \mathbb{R}^n$, an injective differentiable map $A : S \to \mathbb{R}^n$ and a real-valued function $f : A(S) \to \mathbb{R}$,
$$ \int_{y \in A(S)} f(y)\, dy = \int_{x \in S} f(A(x))\, \big| \det(D_A(x)) \big| \, dx $$
So for a linear map $T$,
$$ \int_{y \in T(S)} f(y)\, dy = |\det T| \int_{x \in S} f(T x)\, dx $$
For the map $\Psi^{-1}$ from polar coordinates to rectangular, we get
$$ D_{\Psi^{-1}}(r, \theta) = \left(\begin{array}{cc} \cos \theta & \sin \theta \\
-r \sin \theta & r \cos \theta \end{array}\right) $$
$$ \Big| \det \!\big( D_{\Psi^{-1}}(r, \theta) \big)\Big| = r \cos^2 \theta + r \sin^2 \theta = r $$
giving the familiar
$$ \int_{(x,y) \in S} f(x,y)\, dx\, dy =
\int_{(r,\theta) \in \Psi(S)} f(r \cos\theta, r \sin\theta)\, r\, dr\, d\theta $$
Best Answer
As the others have pointed out, one fundamental use of ordered basis is to specify coordinates. Often, when we say something like $v=(1,2,3)$, we have already assumed that we are talking about coordinates w.r.t. an ordered basis: the canonical basis. If the order is not known, all we know is that $v$ has three coordinates $1,2,3$, but which one is the $x$-coordinate and which ones are $y$ and $z$ would be unknown to us, and hence we cannot be sure where exactly the point $v$ is.
Another purpose of ordering a basis is to make the matrix representation of a linear operator simpler. For instance, consider $f:\mathbb{R}^5\to\mathbb{R}^5$, where the matrix representation of $f$ w.r.t. the canonical basis $A=\{e_1,\ldots,e_5\}$ is given by $$ [f]_A^A=\begin{pmatrix} 3&0&4&0&2\\ 5&4&0&0&4\\ 0&0&2&0&1\\ 2&0&5&1&3\\ 0&0&0&0&1 \end{pmatrix}. $$ Can you tell me the value of $\det f$? For some, the answer can be obtained using only mental calculation, but for the others, the answer is not that easy to obtain. However, if we reorder the basis as $B=\{e_4,e_2,e_1,e_3,e_5\}$, the answer will become utterly obvious: $$ [f]_B^B=\begin{pmatrix} 1&0&2&5&3\\ 0&4&5&0&4\\ 0&0&3&4&2\\ 0&0&0&2&1\\ 0&0&0&0&1 \end{pmatrix}. $$ Admittedly, this example is a bit contrived, but the reordering of bases does sometimes help simplifying the structure of a matrix. For instance, in my answer to a recent question, I had used this trick to make the matrix representation of a certain linear operator block upper triangular, so that I could write down the determinant of this operator immediately.