Jet Nestruev's proof that the exterior derivative $d$ on a real line is not a Kähler differential $d_{C^\infty(\mathbb{R})/\mathbb{R}}$

Question

The relation between the exterior derivative $d:C^\infty(\mathbb{R})\to\Omega^1\mathbb(\mathbb{R})$ and the Kähler differential $d_{C^\infty(\mathbb{R})/\mathbb{R}}:C^\infty(\mathbb{R})\to\Omega_{C^\infty(\mathbb{R})/\mathbb{R}}$ has been already explained in this MO thread. More precisely, the exterior derivative do not satisfy universal property expected from Kähler differentials.

While reading the new edition of Jet Nestruev’s Smooth Manifolds and Observables, I came across section 14.10 which proves the above statement but using simpler tools compared to the MO thread mentioned previously. However, I do not understand a few steps. I will explain the reasoning and point out the steps which I struggle with.

Let $A$ denote the $\mathbb{R}$-algebra of smooth functions on the real line, i.e., $C^\infty(\mathbb{R})$. We want to show that $d_{A/\mathbb{R}}e^x\neq e^xd_{A/\mathbb{R}}x$. The proof goes as follows:

1. We show that $x$ and $e^x$ from $A$ are algebraically independent over $\mathbb{R}$. Thus $\mathbb{R}[x, e^x]\cong \mathbb{R}[X, Y]$ as $\mathbb{R}$-algebras.
2. Since $S:=\mathbb{R}[x, e^x]\setminus \{0\}$ is a multiplicative set of $A$, we can localize $A$ by $S$. Let’s set $A_1:=S^{-1}A$.
3. Additionally, $\mathbb{R}[x, e^x]$ is a domain (as a ring), so that $F:=S^{-1}\mathbb{R}[x, e^x]$ is a field.
4. Let $\delta:=\frac{\partial}{\partial Y}$ (or $\frac{\partial}{\partial e^x}$ if one prefers) be a derivation on the ring $\mathbb{R}[x, e^x]$ induced from $\mathbb{R}[X, Y]$. Using the formula for differentiating fractions, we extend $\delta$ to $F$.
5. Let $I$ be some maximal ideal of $A_1$. Then the field $\bar F:=A_1/I$ contains an isomorphic image of the field $F$, i.e., we have an inclusion $\iota:F\hookrightarrow \bar F$
6. Since $\bar F$ is a field of characteristic zero, the derivation $\delta$ can be extended from $F$ to $\bar F$; let $\bar\delta$ denote that extension.
7. Then $\bar\delta(\bar x)=\overline{\delta(x)}=0$ and $\bar\delta(\overline{e^x})=\overline{\delta(e^x)}=1$, where $\bar{z}=z$ mod $I$ for $z\in A_1$, i.e., $\bar{z}\in A_1/I=\bar{F}$ (see point 5.).
8. Denote the composition $A\to A_1\to\bar{F}\xrightarrow{\bar\delta}\bar{F}$ by $\partial$. It is a $\mathbb{R}$-derivation with values in the $A$-module $\bar{F}$, such that $\partial(x)=0$ and $\partial(e^x)=1$.
9. Due to the universal property of $d_{A/\mathbb{R}}$, there is a $A$-linear homomorphism $h:\Omega_{A/\mathbb{R}}\to\bar{F}$, such that $\partial=h\circ d_{A/\mathbb{R}}$.
10. Since $e^x\partial(x)\neq\partial(e^x)$ (see point 8.), so too $e^xd_{A/\mathbb{R}}x\neq d_{A/\mathbb{R}}e^x$ (see point 9.).

The points 1-4 are easy to follow. In 5, I do not understand why the field $\bar{F}$ must contain an isomorphic image of $F$. In 6, I do not understand why we can extend the derivation from $F$ to $\bar{F}$. Points 7-10 are also clear to me.

In summary, can you give me an explanation of points 5 and 6?