So I am reading Barrett O'Neil's Elementary Differential Geometry, which defines the metric tensor $g_p$ on a surface $M$ to be a bilinear symmetric positive-definite function on the set of ordered pairs of tangent vectors $v,w$ at points $p$ of $M$.
But I read some notes online which uses tensor products, for example, the usual metric on $\mathbb{R}^2$ can be written as $g = dx \otimes dx + dy \otimes dy$, which I'm having trouble understanding.
Theoretically (correct me if I' wrong), I know that the coordinate system $(x_1,x_2)$ gives the basis $(dx_1,dx_2)$ to the cotangent space $T_x^*$, so $g$ which is a bilinear function $T_x \times T_p \to \mathbb{R}$ can be thought of as an element of $T_x^* \otimes T_x^*$, can be represented as a linear combination of $ dx_i \otimes dx_j$.
But how do I show that $ dx \otimes dx + dy \otimes dy$ is the same as the usual dot product on $\mathbb{R}^2$? And how do I make sense of a metric like $ g = \frac{1}{y^2}(dx \otimes dx + dy \otimes dy)?$
The definition of the tensor product is $(f \otimes g)(u,v) = f(u)g(v)$. You should be able to check that whenever $f,g \in T^*$ (i.e. are linear), their product $f \otimes g$ is bilinear; and the product $f \otimes f$ is symmetric and weakly positive definite.
Thus the action of $dx \otimes dx$ on a pair $u,v$ of tangent vectors is $$(dx \otimes dx)(u,v) = dx(u)dx(v) = u_x v_x;$$ so the metric $\delta = dx\otimes dx + dy\otimes dy$ gives $\delta(u,v) = u_x v_x + u_y v_y = u \cdot v$.
In your example $g = \delta/y^2$, this scalar division is simply applied to the output: so $$g(u,v) = \frac{\delta(u,v)}{y^2} = \frac{u\cdot v}{y^2}.$$