Introduction
I do not have a deep understanding of differential forms, and know almost nothing about multiple complex variables. But I think the concepts needed to answer your question won't rely on either; I don't think any major concepts in this exposition will change for $n>1$.
Given any smooth function $\mathbf{f}:\mathbb{R}^{2}\to\mathbb{R}^{2}$,
we may associate $\mathbf{f}\left(x,y\right)=\left\langle u\left(x,y\right),v\left(x,y\right)\right\rangle $
with $g:\mathbb{C}\to\mathbb{C}$ as $g\left(x+iy\right)=u\left(x,y\right)+iv\left(x,y\right)$. I wish I could have two large columns to easily present the parallels
between the real and complex perspectives. With the space limitations
of Mathematics StackExchange, I'm forced to go back and forth.
Cotangent Space
Decomposing a Real Form
We might write something like
\begin{align*}
\mathrm{d}\mathbf{f} & =\left\langle \mathrm{d}u,\mathrm{d}v\right\rangle \\
& =\left\langle \dfrac{\partial u}{\partial x}\mathrm{d}x+\dfrac{\partial u}{\partial y}\mathrm{d}y,\dfrac{\partial v}{\partial x}\mathrm{d}x+\dfrac{\partial v}{\partial y}\mathrm{d}y\right\rangle \\
& =\dfrac{\partial\left\langle u,v\right\rangle }{\partial x}\mathrm{d}x+\dfrac{\partial\left\langle u,v\right\rangle }{\partial y}\mathrm{d}y\\
& =\dfrac{\partial u}{\partial x}\left\langle \mathrm{d}x,0\right\rangle +\dfrac{\partial v}{\partial x}\left\langle 0,\mathrm{d}x\right\rangle +\dfrac{\partial u}{\partial y}\left\langle \mathrm{d}y,0\right\rangle +\dfrac{\partial v}{\partial y}\left\langle 0,\mathrm{d}y\right\rangle \text{.}
\end{align*}
Decomposing a Complex Form
Analogously, we have
\begin{align*}
\mathrm{d}g & =\dfrac{\partial g}{\partial x}\mathrm{d}x+\dfrac{\partial g}{\partial y}\mathrm{d}y\\
& =\dfrac{\partial\left(\mathfrak{R}\circ g+i\mathfrak{I}\circ g\right)}{\partial x}\mathrm{d}x+\dfrac{\partial\left(\mathfrak{R}\circ g+i\mathfrak{I}\circ g\right)}{\partial y}\mathrm{d}y\\
& =\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}+i\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}\right)\mathrm{d}x+\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}+i\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}\right)\mathrm{d}y\\
& =\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}\mathrm{d}x+\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}i\mathrm{d}x+\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\mathrm{d}y+\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}i\mathrm{d}y\text{.}
\end{align*}
Above, I have used $\mathfrak{R}$ and $\mathfrak{I}$ for the real
and imaginary parts.
A New Real Basis
We have written $\mathrm{d}\mathbf{f}$ as a linear combination of
$\left\langle \mathrm{d}x,0\right\rangle $, $\left\langle 0,\mathrm{d}x\right\rangle $,
$\left\langle \mathrm{d}y,0\right\rangle $, $\left\langle 0,\mathrm{d}y\right\rangle $.
And we can rewrite $\mathrm{d}\mathbf{f}$ (or $2\mathrm{d}\mathbf{f}$)
in terms of a new basis: $\left\langle \mathrm{d}x,\mathrm{d}y\right\rangle $,
$\left\langle -\mathrm{d}y,\mathrm{d}x\right\rangle $, $\left\langle \mathrm{d}x,-\mathrm{d}y\right\rangle $,
$\left\langle \mathrm{d}y,\mathrm{d}x\right\rangle $. We have
\begin{align*}
2\mathrm{d}\mathbf{f} & =2\dfrac{\partial u}{\partial x}\left\langle \mathrm{d}x,0\right\rangle +2\dfrac{\partial v}{\partial x}\left\langle 0,\mathrm{d}x\right\rangle +2\dfrac{\partial u}{\partial y}\left\langle \mathrm{d}y,0\right\rangle +2\dfrac{\partial v}{\partial y}\left\langle 0,\mathrm{d}y\right\rangle \\
& =\left(\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}\right)\left\langle \mathrm{d}x,\mathrm{d}y\right\rangle +\left(\dfrac{\partial v}{\partial x}-\dfrac{\partial u}{\partial y}\right)\left\langle -\mathrm{d}y,\mathrm{d}x\right\rangle \\
& \phantom{=}+\left(\dfrac{\partial u}{\partial x}-\dfrac{\partial v}{\partial y}\right)\left\langle \mathrm{d}x,-\mathrm{d}y\right\rangle +\left(\dfrac{\partial v}{\partial x}+\dfrac{\partial u}{\partial y}\right)\left\langle \mathrm{d}y,\mathrm{d}x\right\rangle \text{.}
\end{align*}
As an aside, note that the latter two coefficients would be $0$ if
the Cauchy-Riemann equations
are satisfied.
A New Complex Basis
We have written $\mathrm{d}g$ as a real linear combination
of $\mathrm{d}x$, $i\mathrm{d}x$, $\mathrm{d}y$, $i\mathrm{d}y$.
And we can rewrite $\mathrm{d}g$ (or $2\mathrm{d}g$) in terms of
a new basis: $\mathrm{d}x+i\mathrm{d}y$, $-\mathrm{d}y+i\mathrm{d}x$,
$\mathrm{d}x-i\mathrm{d}y$, $\mathrm{d}y+i\mathrm{d}x$. Note that
those expressions can be rewritten in a suggestive way as $\mathrm{d}x+i\mathrm{d}y$,
$i\left(\mathrm{d}x+i\mathrm{d}y\right)$, $\mathrm{d}x-i\mathrm{d}y$,
$i\left(\mathrm{d}x-i\mathrm{d}y\right)$. Inspired by $z=x+iy$ and $\overline{z}=x-iy$, we can make this even tidier by introducing the abbreviations $\mathrm{d}z=\mathrm{d}x+i\mathrm{d}y$ and $\mathrm{d}\overline{z}=\mathrm{d}x-i\mathrm{d}y$;
this reduces the basis to $\mathrm{d}z$, $i\mathrm{d}z$, $\mathrm{d}\overline{z}$,
$i\mathrm{d}\overline{z}$. We have
\begin{align*}
2\mathrm{d}g & =2\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}\mathrm{d}x+2\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}i\mathrm{d}x+2\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\mathrm{d}y+2\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}i\mathrm{d}y\\
& =\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}+\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}\right)\left(\mathrm{d}x+i\mathrm{d}y\right)+\left(\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}-\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\right)\left(-\mathrm{d}y+i\mathrm{d}x\right)\\
& \phantom{=}+\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}-\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}\right)\left(\mathrm{d}x-i\mathrm{d}y\right)+\left(\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}+\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\right)\left(\mathrm{d}y+i\mathrm{d}x\right)\\
& =\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}+\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}\right)\mathrm{d}z+\left(\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}-\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\right)\left(i\mathrm{d}z\right)\\
& \phantom{=}+\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}-\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}\right)\mathrm{d}\overline{z}+\left(\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}+\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\right)\left(i\mathrm{d}\overline{z}\right)\\
& =\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}+i\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}+\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}-i\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\right)\mathrm{d}z\\
& \phantom{=}+\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}+i\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}-\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}+i\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\right)\mathrm{d}\overline{z}\\
& =\left(\dfrac{\partial\left(\mathfrak{R}\circ g+i\mathfrak{I}\circ g\right)}{\partial x}+\dfrac{\partial\left(\mathfrak{I}\circ g-i\mathfrak{R}\circ g\right)}{\partial y}\right)\mathrm{d}z\\
& \phantom{=}+\left(\dfrac{\partial\left(\mathfrak{R}\circ g+i\mathfrak{I}\circ g\right)}{\partial x}+\dfrac{\partial\left(i\mathfrak{R}\circ g-\mathfrak{I}\circ g\right)}{\partial y}\right)\mathrm{d}\overline{z}\\
& =\left(\dfrac{\partial g}{\partial x}+\dfrac{\partial\left(-ig\right)}{\partial y}\right)\mathrm{d}z+\left(\dfrac{\partial g}{\partial x}+\dfrac{\partial\left(ig\right)}{\partial y}\right)\mathrm{d}\overline{z}\\
& =\left(\dfrac{\partial}{\partial x}-i\dfrac{\partial}{\partial y}\right)g\,\mathrm{d}z+\left(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\right)g\,\mathrm{d}\overline{z}\text{.}
\end{align*}
Note that $\mathrm{d}g=\dfrac{\partial g}{\partial x}\mathrm{d}x+\dfrac{\partial g}{\partial y}\mathrm{d}y$
started as a complex linear combination of $\mathrm{d}x$ and $\mathrm{d}y$,
and ended up as a complex linear combination of $\mathrm{d}z$ and
$\mathrm{d}\overline{z}$, even though we passed through a representation
as a real linear combination to show the parallel with $\mathbf{f}$.
Dividing through by $2$, we obtain $\mathrm{d}g=\dfrac{1}{2}\left(\dfrac{\partial}{\partial x}-i\dfrac{\partial}{\partial y}\right)g\,\mathrm{d}z+\dfrac{1}{2}\left(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\right)g\,\mathrm{d}\overline{z}$.
This form suggests defining the "Wirtinger derivatives"
: $\dfrac{\partial}{\partial z}=\dfrac{1}{2}\left(\dfrac{\partial}{\partial x}-i\dfrac{\partial}{\partial y}\right)$
and $\dfrac{\partial}{\partial\overline{z}}=\dfrac{1}{2}\left(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\right)$
so that we may write the tidy $\boxed{\mathrm{d}g=\dfrac{\partial g}{\partial z}\mathrm{d}z+\dfrac{\partial g}{\partial\overline{z}}\mathrm{d}\overline{z}}$.
In this form, the Cauchy-Riemann equations amount to "$\dfrac{\partial g}{\partial\overline{z}}=0$ as a complex number".
Cauchy Integral Formula
As you noted, the Cauchy Integral Formula doesn't usually have that extra term, because it's not needed for holomorphic functions. This generalization for smooth functions is sometimes known as the Cauchy-Pompeiu formula. An exposition (starting from an understanding of the boxed equation above) can be found in C.K. Fong's notes for "Calculus of Differential Forms with Applications". Specifically, in "Section 2. Complex 2 - forms: Cauchy - Pompeiu's Formula" of "Chapter 3: Complex Differential
Forms". I'll summarize the argument here, translated into your book's variables.
Let $z$ be a particular complex number, and use $w$ for a varying
one. Consider $\dfrac{f\left(w\right)}{w-z}\mathrm{d}w$. Then a routine
calculation (using $\mathrm{d}\left(\dfrac{\mathrm{d}w}{w-z}\right)=0$)
shows that $\mathrm{d}\left(\dfrac{f\left(w\right)}{w-z}\mathrm{d}w\right)=\dfrac{\partial f(w)}{\partial\overline{w}}\dfrac{\mathrm{d}\overline{w}\wedge\mathrm{d}w}{w-z}$.
We can apply Green's Theorem/general Stokes' Theorem to the large disc
$\Delta$ by removing a small disc around the point
$z$ and letting the small radius go to zero, just as in the standard proof of the Cauchy Integral Formula
for a holomorphic $f$.
By the standard argument about what happens
to the integral of $\dfrac{f(w)\mathrm{d}w}{w-z}$ around a circle
whose radius approaches $0$, we end up with $\iint_{\Delta}\dfrac{\partial f(w)}{\partial\overline{w}}\dfrac{\mathrm{d}\overline{w}\wedge\mathrm{d}w}{w-z}=\int_{\partial\Delta}\dfrac{f(w)\mathrm{d}w}{w-z}-2\pi if\left(z\right)$.
This is equivalent to the formula in your book (make sure to note
that $\mathrm{d}\overline{w}\wedge\mathrm{d}w=-\mathrm{d}w\wedge\mathrm{d}\overline{w}$).
Note that only when $f$ is holomorphic could you note that $\dfrac{\partial f(w)}{\partial\overline{w}}=0$ and simplify.
