Inner product between tangent vectors Suppose we are given a point $p$ on a Riemannian manifold $M$ with
a tangent space $V=T_{p}M$. Furthur assume that this tangent space
is Euclidean. Suppose we have two tangent vectors represented in differential
geometry as vectors $\mathbf{A}=A^{i}\partial_{i}$and $\mathbf{B}=B^{j}\partial_{j}.$
My question is what would the dot product between two such vectors
($V\times V\rightarrow\mathbb{R})$ look like? In an orthogonal coordinate
system, how could we show that $<\partial_{i},\partial_{j}>=\delta_{ij}$?
Note that I am intentionally avoiding one-forms or covectors here,
since in linear algebra the scalar product is defined on $V\times V\rightarrow\mathbb{R}$.
I know how to form a scalar product between a vector and covector
so that's not at issue.
 A: The tangent space $V = T_pM$ has been endowed with a metric, a bilinear form $g$. Given a basis $\{v_i\}$ for $V$, the metric has coordinates $g_{ij} = g(v_i,v_j)$. In general, the dot product between two vectors $A$ and $B$ is given by $g(A,B)$; in coordinates, with $A = \sum A^iv_i$ and $B = \sum B^jv_j$, then the dot product is 
$$ g(A,B) = \sum_{i,j} g_{ij}A^iB^j.$$
(You can prove this quickly by appealing to bilinearity.)
If you choose an orthonormal basis for $T_pM$, then by definition of orthonormality the coordinates of $g$ are $g_{ij} = \delta_{ij}$. An orthonormal basis always exists, and can be constructed via the Gram-Schmidt process.
Beware: In general an orthonormal basis of $V$ does not come from a local coordinate system on the manifold $M$. That is, for coordinate vector fields, you should not assume that $g(\partial_i,\partial_j)=\delta_{ij}$. The obstruction for this happening in a neighborhood is the curvature tensor.
To elaborate, 


*

*Coordinate functions $x^i:U\to\mathbb{R}$ induce a basis $\frac{\partial}{\partial x^i}$ at each point in the coordinate patch. These induce coordinate functions $g_{ij}$ for the metric. Unless curvature vanishes, $g_{ij}$ will not be equal to $\delta_{ij}$ in any open subset of the patch.

*It is possible to choose a set of vector fields $\{e_i\}$ on an open set of the manifold such that at each point the $\{e_i\}$ form an orthonormal basis of the tangent space. In these bases, the metric will at each point of this open set be $\delta_{ij}$. However, unless curvature vanishes, the orthonormal framing will not be integrable to coordinates. (As above, an orthonormal frame field can be created by applying Gram-Schmidt pointwise to coordinate vector fields.)

*It is possible to choose coordinates so that at a single point the coordinate functions of the metric are $\delta_{ij}$. One way of doing this is normal coordinates. However, unless curvature vanishes, away from that point, in those coordinates the metric will no longer be the identity matrix.


Addendum
An inner product is a choice of bilinear form $g:V\times V\to \mathbb{R}$. That is, once an inner product has been fixed, you already know how to multiply two tangent vectors. What I explain above is how to find the expression for this multiplication in coordinates on $V$. 
Once an inner product has been chosen, then you can define the musical isomorphism via 
$$ v^\flat(w) = g(v,w). $$
So in the sense that the musical isomorphism is defined exactly so that $v^\flat(w) = g(v,w)$, finding $g(v,w)$ involves finding $v^\flat$ and then contracting with $w$, but "morally" the definition of $g$ precedes the definition of $\cdot^\flat$.

For further reading, I recommend O'Neill, Semi-Riemannian Geometry with an Introduction to Relativity. Chapters 1 and 2 cover the basic smooth manifold construction and multilinear algebra that you're asking about here. This may be especially suited to you because I'm guessing you come from a physics background.
Other introductory books to differential geometry will also cover this material at varying levels of abstraction. Do Carmo's book Riemannian Geometry is a standard intro graduate text. Warner's book Introduction to Differentiable Manifolds covers the same material in the first couple of chapters, but more abstractly.
