Can we define a specific metric tensor to all manifolds? First, my question it maybe doesn't make sense at all, and that's because I don't truly understand these things, if it doesn't make sense please explain to me why.
When I say a metric tensor I mean the first fundamental form where every student learns in a differential geometry course. Meaning the standard inner product $\|w\|=\langle w,w\rangle $ of an element of the tangent space $T_pS$ of a normal surface $S$. Can we use this for every n-manifold to calculate length, angles, surface area etc?
what's the point of considering all these other tensors we use in Riemannian geometry since a manifold is locally euclidean can't we just use the standard inner product to do all these things?
 A: (1) When most people say "an $n$-manifold $M^n$," what they mean typically mean is an (abstract) smooth $n$-manifold (which is virtually the same as an (abstract) differentiable $n$-manifold).  I don't have time to go through the definition here, but there are many fantastic books on abstract manifolds (e.g., John Lee's Introduction to Smooth Manifolds), as well as a Wikipedia page:
https://en.wikipedia.org/wiki/Differentiable_manifold
The point is that an abstract $n$-manifold $M^n$ is not necessarily a subset of euclidean space $\mathbb{R}^N$.  However, if it happens to be the case that your abstract $n$-manifold lives in some euclidean space, say $M^n \subset \mathbb{R}^N$, then we say that $M^n$ is a smooth submanifold of $\mathbb{R}^N$. If you've taken a differential geometry course on surfaces in $\mathbb{R}^3$, then you've studied the smooth $2$-dimensional submanifolds of $\mathbb{R}^3$.
(2) Given a smooth submanifold $M^n \subset \mathbb{R}^N$, then yes, it is possible to use the standard inner product $\Vert v \Vert = \sqrt{\langle v,v \rangle}$ on $\mathbb{R}^N$ to give each tangent space $T_pM \subset \mathbb{R}^N$ a positive-definite inner product that varies smoothly with $p \in M$:
\begin{equation}
g_p(v,w) := \langle v,w \rangle.
\tag{1}
\end{equation}
(3) In general, however, an abstract manifold $M^n$ doesn't necessarily lie in a euclidean space $\mathbb{R}^N$.  Nevertheless, it would be interesting if we could give each tangent space $T_pM$ a smoothly varying positive-definite inner product $g_p$.  A choice of smoothly varying positive-definite inner product on each $T_pM$ is called a Riemannian metric (or metric tensor or Riemannian structure).  The pair $(M,g)$ is then called a Riemannian manifold.
It is a fact (mentioned in Moe's answer) that every abstract $n$-manifold can be given (uncountably many!) Riemannian metrics $g$.  Therefore, every abstract manifold $M$ can be made into a Riemannian manifold $(M,g)$ in uncountably many ways.  If it so happens that $M^n \subset \mathbb{R}^N$, then there is a natural choice of Riemannian metric $g$ coming from equation (1), but one could in principle give $M^n$ other Riemannian metrics (a.k.a. metric tensors) besides this.
A: $\newcommand{\R}{\mathbb{R}}$
Even though you're asking about manifolds in general, it suffices to understand this when the manifold is just an open set $M \subset \mathbb{R}^n$. The fact that it is a subset of $\R^n$ means there is already a set of coordinate functions $x^k: M \rightarrow \R$. However, there can be other coordinate functions $y^k: M \rightarrow \R$. For example, if $M = (0,1) \times (0,2\pi)$, the standard coordinates would be $x^1(a,b) = a$ and $x^2(a,b) = b$. Another set of coordinates would be $y^1(a,b) = a\cos b$ and $y^2 = a\sin b$.
When studying geometry or physics, coordinates are a necessary evil because we need to be able to measure and compare things. However, geometric properties and physical laws should not depend on the coordinates used. For example, physical laws should not depend on whether we are measuring lengths in meters or feet.
The first step towards defining a Riemannian metric is understanding what a tangent space is. The key idea here is that every point $p \in M$ has its own tangent space $T_pM$. It consists of all possible velocity vectors of curves. If we fix one set of coordinates, then $T_pM$ is obviously isomorphic to $\R^n$. If we switch to a different set of coordinates, then $T_pM$ still isomorphic to $\R^n$, but the isomorphism is different from the first one. The crucial observation is that the map $\R^n \rightarrow T_pM \rightarrow \R^n$ defined using the two different sets of coordinates is linear. What this allows us to do is to view $T_pM$ as an abstract vector space, where a set of coordinates defines an isomorphism $T_pM \rightarrow \R^n$. Put differently, a set of coordinates implies a basis of $T_pM$, commonly denoted as $(\partial_1, \dots, \partial_n)$.
Now recall that a dot or inner product on an abstract vector space $V$ is a bilinear function $g: V \times V \rightarrow \R$ such that $g(v,v) \ge 0$, with equality holding only if $v = 0$. A Riemannian metric is simply an inner product defined on each $T_pM$. It is important to note that we are not assuming any relationship between the inner products on two different tangent spaces $T_pM$ and $T_qM$.
Next, recall that, given a basis $(v_1, \dots, v_n)$ of $V$, an inner product $g$
is uniquely determined by the positive definite symmetric matrix $[g_{ij}]$, where
$$ g_{ij} = g(v_i,v_u).$$
So if we choose a set of coordinates, $(x^1, \dots, x^n)$ on $M$, then, at each point, we have a basis $(\partial_1, \dots, \partial_n)$ of $T_pM$, which then allows us to write the Riemannian metric at $p$ as the matrix $[g_{ij}(p)]$, where
$$
g_{ij}(p) = g(p)(\partial_i, \partial_j)
$$
Here's the next important point: This matrix is a matrix of functions that depend on the point on the manifold. On top of that, the matrix depends on which coordinates you used. What's important is that the matrix changes in just the right way, so that, no matter what coordinates you use, the matrix, used with the basis of $T_pM$ defined by the coordinates, defines the same abstract inner product on $T_pM$.
To study this further, I encourage you to look at fundamental examples such as the standard sphere, hyperbolic space, surfaces of revolution in $\R^3$, and graphs of functions in $\R^n$.
A: Every differentiable manifold admits a Riemannian structure. This is actually fairly intuitive to see: Every differentiable manifold is locally Riemannian, thus sewing up all the local metrics to a global metric gives us the conclusion we seek.
