Euclidean metric on a Riemannian manifold Lets say we have a Euclidean configurations space $\mathbb E^n$ equipped with a smooth inner product $\langle \cdot ,\cdot \rangle$ with positive signature in the tangent space above each point. We have defined a Riemannian manifold. 
We can also call this inner product a metric tensor $g$, such that if $g$ acts on two vectors then $g(v,w)$ where $v,w\in T_p\mathbb E^n$ (tangent space to a point $p$).
From general googling and piecing things together I am lead to write another expression for $g$, namely,
\begin{equation}
\boxed{
g(v,v)=g_{ij}\frac{dx^i}{dt}\frac{dx^j}{dt}} 
\end{equation}
Where, 
\begin{equation}
v=v^ie_i
\end{equation}
Is this something we can do? My reasoning is that $\|v\|=\sqrt{v\cdot  v}=\sqrt{g(v,v)}$ from wikipedia and I have seen (page 5),
\begin{equation}
\|v\|=\sqrt{g_{ij}\frac{dx}{dt}\frac{dx}{dt}}
\end{equation}
Therefore is the boxed expression above correct? In addition I am assuming $v=\dot x$. If this is so then I assume that $v$ would have to be the representative of $x$ in $T_p\mathbb E^n$?
For physical application I am trying to understand how the following simple Lagrangian is constructed, 
 \begin{equation}
\mathscr L=\frac{1}{2}\sum _{ij}\text{g}_{ij}\dot q^i\dot q^j
\end{equation}
I would just add that I am not very confident with tensorial notation, while I realise that on the surface this may look correct I feel I may be trading over important details? 
 A: Your boxed equation is that of the line element, not the metric tensor. 
For a line element, we can write 
$ds^2 = g_{ab} v^a v^b d\lambda^2$, 
where we have parametrized some curve by $\lambda$ and $v^a$ are tangent vectors. 
If we set
$v^a = \frac{dx^a}{d\lambda}$ 
then we obtain the standard line element expression: 
$ds^2 = g_{ab} dx^a dx^b$. 
Recall that the metric tensor is a generalization of the scalar product. What you wrote was an expression of the line element. 
UPDATE:
What you end up constructing is the Lagrangian of geodesics. That is, consider the Lagrangian function along a path $\Gamma$ described by $x^{m}(\lambda)$, where once again we parameterize the curve $\Gamma$ by $\lambda$. Now, we simply take the Lagrangian function to be: $L = \left(\frac{ds}{d\lambda}\right)^2$, so that for a line element $ds^2 = g_{mn}dx^{m}dx^{n}$, we may write: $L = g_{mn}\dot{x}^{m} \dot{x}^{n}$. You have written this with a $(1/2)$ factor, but, this is usually done for convenience, as I will show below.
The point is once you obtain this Lagrangian, you wish to derive conditions on that path that extremizes:
$\int L d\lambda$.
This is just given by the Euler-Lagrange equations:
$\frac{d}{d\lambda} \left(\frac{\partial L}{\partial \dot{x}^{c}}\right) = \frac{\partial L}{\partial x^{c}}$. 
In fact, one can show that the Euler-Lagrange equations are precisely the geodesic equations:
$\ddot{x}^{e} + \Gamma^{e}_{mb}\dot{x}^{m} \dot{x}^{b} = 0$,
where $\Gamma^{e}_{mb}$ are the standard Christoffel symbols derived from the metric tensor.
Now for the fun part. I used a Lagrangian as is commonly done in G.R.:
$L = g_{mn}\dot{x}^{m} \dot{x}^{n}$, 
you stated:
$L = \frac{1}{2}g_{mn}\dot{x}^{m} \dot{x}^{n}$, 
but does it really matter? 
Let us consider the Euler-Lagrange equations for the more general Lagrangian $F(L)$, where $F$ is now any function of $L$ along the path $\Gamma$. 
First, note that
$\frac{\partial F}{\partial x^{c}} = \frac{dF}{dL} \frac{\partial L}{\partial x^{c}}$, 
and
$\frac{\partial F}{\partial \dot{x}^{c}} = \frac{dF}{dL} \frac{\partial L}{\partial \dot{x}^{c}}$.
Now, we have that:
$\frac{d}{d\lambda} \left(\frac{\partial F}{\partial \dot{x}^{c}}\right) - \frac{\partial F}{\partial x^{c}} = \frac{dF}{dL} \left[\frac{d}{d\lambda} \left(\frac{\partial L}{\partial \dot{x}^{c}}\right) - \frac{\partial L}{\partial x^{c}}\right] = 0$,
where the last expression in brackets is precisely the Euler-Lagrange equations. So, we can see that our original choice for $L$ will also extremize any $F(L)$ as well along the path. 
A: The expression you are asking about (correctly) defines the metric on the diagonal, i.e. on pairs of the form $(v,v)$. This actually is all you need by bilinearity and symmetry: if you know $g(v,v)$, $g(w,w)$ and $g(v+w,v+w)$ you also know $g(v,w)$. Note that what you are really saying is that $v$ is a tangent vector at some point $P$ that is realized as the tangent vector to a parametrized curve (with parameter $t$) whose component functions in your fixed local coordinates are denoted $x^i(t)$.
The Lagrangian you write down is sometimes called the energy functional and its Euler-Lagrange equations are exactly the geodesic equations, i.e. action minimizing curves for this Lagrangian are geodesics, curves that locally realize minimal distance and are parametrized at constant velocity.
