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?
(1) When most people say &quot;an $n$-manifold $M^n$,&quot; 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.
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.