What's the precise definition of coordinates in Euclidean space?

About convention/notation:

What does the expression $$\text{Let }(z^1,z^2,\dots,z^n)\text{ denote the coordinates on $V$}$$ mean?

The formalized version of this expression is

If $(V,\varphi)$ is a chart on some $n$-dimensional manifold then let $$(z^1,z^2,\dots,z^n) = (\pi_1\circ\varphi,\pi_2\circ\varphi,\dots,\pi_n\circ\varphi)$$ where $\pi_i$ is the $i^{th}$ projection on $\mathbb{R}^n$. That is, $z^i : V \rightarrow \mathbb{R}$ is defined by $$z^i(p) = \pi_i(\varphi(p))$$ for all $p \in V$.

Note that the expression references, tacitly, some ambient manifold and a chart (which is usually clear from context).

Of course, an equivalent version of this is defined in your post, but I'm stating it here in order to fully address your concern about the Corollary. Also, the above expression is often used where the manifold is smooth (and so is the chart).

About Partial Derivatives:

Firstly,

Partial derivatives are well-defined on any smooth chart $(V,\varphi)$ belonging to a smooth manifold.

Here is how it is defined:

Let $M$ denote an $n$-dimensional smooth manifold, and $(V,\varphi)$ a smooth chart on $M$. Let $(z^1,z^2,\dots,z^n)$ denote the coordinates on $V$. If $f \in C^\infty(V)$, then the $i^{th}$ partial derivative operator $\frac{\partial f}{\partial z^i} : V \rightarrow \mathbb{R}$ defined at $f$ is $$\frac{\partial f}{\partial z^i}(p) = \frac{\partial (f\circ \varphi^{-1})}{\partial x^i}(\varphi(p))$$ for $p \in V$, where the second expression uses the regular $i^{th}$ partial derivative operator in $\mathbb{R}^n$.

I left out a lot of details, as well as the path toward motivation for its definition, for brevity (and focus) of this post. To see these details and why the above is well-defined, see Chapter 3, "Tangent Vectors", of the text (see the section "Computations in Coordinates" in particular).

Note that in the above, you're not "differentiating with respect to" $z^i$ (which is senseless in this context as you've mentioned), you're just using $z^i$ in the notation to reference the chart you're defining the operator with.

This is helpful for your question because in the case of the Corollary, the expression

Let $(x^1,x^2,\dots,x^n)$ denote the standard coordinates on $U \subseteq \mathbb{R}^n$.

means

Let $(x^1,x^2,\dots,x^n)$ denote the coordinates on $U$ (see the section About convention/notation above) where the ambient manifold is $\mathbb{R}^n$ with its standard smooth structure and the smooth chart under consideration is $(U,id_U)$.

so that if $f \in C^\infty(U)$ then using the above definition of partial derivatives we have that for $x \in U$,

$$\frac{\partial f}{\partial x^i}(x) = \frac{\partial (f \circ id_U^{-1})}{\partial x^i}(id_U(x)) = \frac{\partial f}{\partial x^i}(x)$$

That is, in the case where $U$ is given standard coordinates, the operation is the same as the regular partial derivative operator. The $x^i$ on the left corresponds to coordinates on $U$, and the $x^i$ on the right corresponds to the notation used for the definition of the regular partial derivative operator in $\mathbb{R}^n$. For further clarity,

If $(z^1,z^2,\dots,z^n)$ denotes the standard coordinates on $U \subseteq \mathbb{R}^n$, then $$ \frac{\partial f}{\partial z^i}(x) = \frac{\partial (f \circ id_U^{-1})}{\partial x^i}(id_U(x)) = \frac{\partial f}{\partial x^i}(x)$$

Thus, for the case of the notation within the Corollary,

$$\frac{\partial G^i}{\partial y^k}(F(x)) = \frac{\partial (G_i \circ id_\widetilde{U}^{-1})}{\partial x^k}(id_\widetilde{U}(F(x))) = \frac{\partial G_i}{\partial x^k}(F(x))$$

where the $x^k$ on the right corresponds to the notation used for the regular $k^{th}$ partial derivative in $\mathbb{R}^m$.

Caveat:

This is a book about smooth manifolds, so the above partial derivative operator is only defined for cases where $f \in C^\infty(U)$ rather than $f \in C^1(U)$ as required within the Corollary. Of course we know the Corollary to be true from classical analysis, and it can be proved without any direct mention of ideas from the theory of smooth manifolds. The main reason I believe why the author mentioned the notation/convention is to differentiate between the regular partial derivative operator within $\mathbb{R}^n$ and $\mathbb{R}^m$ (since the book only uses the $x^i$ notation to define the regular partial derivative, a scattering of only $x^i$'s would've made the Corollary unpleasant to read) as well as to begin to bring in some of the convention/notation that will be used within the text.

You mention bases in linear algebra - this is very, very similar. When you write $\mathbb{R}^n$ here, you mean an $n$-dimensional $\mathbb{R}$-vector space $V$ with a chosen ordered basis (you could be picky and just think about the set of $n$-tuples of real numbers with the usual addition and scalar multiplication and the ordered basis of tuples like $(1,0,\dots,0)$ that you'd likely choose); let's say $b_1,\dots,b_n \in V$ is this ordered basis. There is a unique ordered basis (the dual basis) in the dual of our vector space $V^*$, which we'll denote by $x^1,\dots,x^n \in V^*$, so that $$x^i(b_j) = \left\{ \begin{array}{ll} 1 & i=j \\ 0 & i\ne j \end{array} \right.$$ for any $1\le i,j\le n$. By definition, these $x^i$ are (linear) functions from our vector space $V$ (though we abuse notation and just write $\mathbb{R}^n$ instead of $V$) to the underlying field $\mathbb{R}$. This is all cooked in when you write $\mathbb{R}^n$. These functions $x^1,\dots,x^n$ are the standard coordinates on $\mathbb{R}^n$.

Now if you take any open subset $U\subset \mathbb{R}^n$, you can just restrict these functions $x^1,\dots,x^n$ to $U$ and call them the standard coordinates on $U$ (even though $U$ is likely no longer a vector space).