Deriving the round metric on $\mathbb{S}^2$

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Consider $\mathbb{S}^2$ as a submanifold of $\mathbb{R}^{3}$. Given that $\mathbb{R}^{3}$ has the standard Riemannian metric $g$, I want to derive a local expression or the metric tensor for the round metric $\mathring{g}$ on $\mathbb{S}^2$.

This is given by $\mathring{g}=\iota^*g$ so for $u,v \in T_p\mathbb{S}^2$ one has $$\mathring{g}(u,v)=g_{\iota(p)}(d\iota_p(u),d\iota_p(v)).$$

Also in coordiantes we can express $$g=\sum g_{ij}dx^i \otimes dx^j$$ where $g_{ij}=g_p(\partial/\partial x^i,\partial/\partial x^j)$ so the pullback would result in $$\mathring{g}=\sum (\iota^*g)_{ij} dx^i \otimes dx^j.$$ To find the local expression I need to compute $$(\iota^*g)_{ij}$$ but what is this expression? I don't know how to compute it. Do I need to parameterize $\mathbb{S}^2$ or can I do this without it?

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Your formula for the pull-back is not right. Pulling-back a tensor is a linear operation. Moreover, as mentioned in @jd27's comment, $dx^i$ are not well-defined on $S^2$ as it stands.

Formally, the pull-back of $g$ by $\iota$ should read \begin{align} \iota^*g &= \iota^*\left( \sum_{i,j=1}^3g_{ij} dx^i\otimes dx^j \right) \\ &= \sum_{i,j=1}^3 \iota^*\left(g_{ij}dx^i\otimes dx^j \right) \\ &= \sum_{i,j=1}^3 (g_{ij}\circ \iota)\, \iota^*( dx^i \otimes dx^j) \\ &= \sum_{i,j=1}^3 (g_{ij}\circ \iota)\, \iota^*(dx^i) \otimes \iota^*(dx^j) \\ &= \sum_{i,j=1}^3 (g_{ij}\circ \iota)\, d(x^i\circ \iota) \otimes d(x^j\circ \iota). \end{align}

But then, this is not an expression of the round metric in coordinates for the sphere. Indeed, $\{x^1\circ \iota, x^2\circ \iota,x^3\circ \iota\}$ is not a coordinate patch on the sphere, for the simple reason that the sphere is two dimensional, so any coordinate patch should have only two coordinate functions.

To derive a coordinate expression of the sphere, you need to first consider a coordinate patch $\{y^1,y^2\}$ on $S^2$ and then look at $\iota^*g$ in these coordinates. There are as many expressions as coordinate patches: an infinite number of them. Usually, one would use polar coordinates or stereographical ones.