Why $\partial$ and $\bar{\partial}$ defined in that way (the Wirtinger derivatives)?

If $f:\mathbb{C}\to\mathbb{C}$ is any smooth function and $z\in\mathbb{C}$, the derivative $df_z$ of $f$ at $z$ is a $\mathbb{R}$-linear operator from the tangent space $T_{z}\mathbb{C}$ to $\mathbb{C}$ (and $T_{z}\mathbb{C}$ can of course be canonically identified with $\mathbb{C}$ since $\mathbb{C}$ is a vector space). Said differently, $df$ is a complex-valued differential $1$-form on $\mathbb{C}$ (i.e. an element of the space $\Omega^1(\mathbb{C};\mathbb{C})=\Omega^1(\mathbb{C})\otimes_{\mathbb{R}} \mathbb{C}$).

Any $\mathbb{R}$-linear operator $A$ from one complex vector space to another can be written uniquely as $A=B+C$, where $B$ is complex-linear (i.e. $B(iv)=iBv$ for all $v$) and $C$ is complex anti-linear (i.e. $B(iv)=-iBv$). Specifically, let $Bv=\frac{1}{2}(Av-iAiv)$ and $Cv=\frac{1}{2}(Av+iAiv)$.

The operators $\partial$ and $\bar{\partial}$ can then be characterized as follows: for a function $f:\mathbb{C}\to\mathbb{C}$, one has the unique decomposition $$ df=\partial f+\bar{\partial} f $$ where the complex-valued $1$-forms $\partial f$ and $\bar{\partial}f$ are such that, at each $z\in\mathbb{C}$, $(\partial f)_z$ is complex-linear and $(\bar{\partial} f)_z$ is complex anti-linear. At least to me that justifies the notation--it makes more sense to have $\partial f$ be the linear part of $df$ and $\bar{\partial} f$ be the antilinear part than the other way around.

As for $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$, note first that as a special case of the above discussion one has functions $z:\mathbb{C}\to\mathbb{C}$ (which is just the identity) and $\bar{z}:\mathbb{C}\to\mathbb{C}$, and therefore complexified one-forms $dz,d\bar{z}\in\Omega^1(\mathbb{C};\mathbb{C})$. (More specifically, since $z=x+iy$ and $\bar{z}=x-iy$, one has $dz=dx+idy$ and $d\bar{z}=dx-idy$, so for $dz$ and $d\bar{z}$ the signs on $i$ are what you wanted them to be.)

At each point of $\mathbb{C}$, $\{dz,d\bar{z}\}$ is a basis (over $\mathbb{C}$) for the complexified cotangent space. The operators $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$ are then naturally defined as the complexified vector fields such that at every point the basis $\{\frac{\partial}{\partial z},\frac{\partial}{\partial\bar{z}}\}$ for the complexified tangent space is the dual basis to the basis $\{dz,d\bar{z}\}$ for the complexified cotangent space. Imposing this dual basis requirement, the formulas $dz=dx+idy$ and $d\bar{z}=dx-idy$ then readily yield the standard formulas for $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$, and give rise to the pleasant identity, for any smooth $f:\mathbb{C}\to\mathbb{C}$, $$ df=\partial f+\bar{\partial} f=\frac{\partial f}{\partial z}dz+\frac{\partial f}{\partial{\bar{z}}}d\bar{z} $$ just like one would get if $z$ and $\bar{z}$ were real coordinates on a real two-manifold.

Explaining this notation may be as thankless a task as explaining a joke, but here goes.

It may help to think in terms of power series. Consider power series in two variables $x$ and $y$ with complex coefficients, instead of complex functions of real variables $x$ and $y$. Now note that such a thing can be rewritten as a power series in variables $z$ and $\bar z$ by the substitutions $x=\frac{1}{2}(z+\bar z)$ and $y=\frac{1}{2i}(z-\bar z)$. And you can convert back to the other form by the substitutions $z=x+iy$ and $\bar z=x-iy$. The holomorphic case is the case where the only terms in the $(z,\bar z)$ power series are the powers of $z$.

Whether you think in terms of power series or not, there is a fruitful fiction that complex functions of $x$ and $y$ can be thought of alternatively as functions of $z$ and $\bar z$, with the holomorphic ones being the functions of $z$ alone. This is consistent with the notation that you are asking about; see Qiaochu's comment to the question.

When I was a grad student the story went around that a fellow student attending the complex analysis course questioned precisely these signs, persistently suggesting that the professor had got them wrong. Finally the professor responded "Ah, I see! You and I must be thinking of different square roots of $-1$!"

(Edit: That's a joke.)

This same fellow student, when we were planning a skit for the annual math department picnic, objected to the title "Let Sleeping Dilogs" on the grounds that people who didn't know about dilogarithms wouldn't get the joke. He wanted to call it "Let Sleeping Dogs Lie" instead, which would have meant that there was no joke to get.

This is actually a question of linear algebra. The apparently inappropriate choice of sign stems from the fact that if $W$ is a (complex) vector space, and $A$ an endomorphism of $W$, then $$ \ker (A-\lambda 1)\subset W $$ is the eigenspace for eigenvalue $(+\lambda)$.

In our case, we start with a real vector space, $V$, and an endomorphism $I$, satisfying $I^2=-1$. We extend it $\mathbb{C}$-linearly to an endomorphism $I_{\mathbb{C}}$ of $V\otimes \mathbb{C}$, and decompose the latter into $\pm i$ eigenspaces, usually called $V^{1,0}$ and $V^{0,1}$. The identity transformation decomposes into a sum of (eigenspace) projectors: $$ 1 = \frac{1}{2}\left( 1 - i I_{\mathbb{C}}\right) + \frac{1}{2}\left( 1 + i I_{\mathbb{C}}\right),$$ and we have a $\mathbb{C}$-linear isomorphism $V\subset V\otimes\mathbb{C}\to V^{1,0} = \ker\frac{1}{2}\left( 1 + i I_{\mathbb{C}}\right) $. Under this isomorphism $V\ni v\mapsto \frac{1}{2}(v-iIv)\in V^{1,0}$.

The complex structure $I$ induces a complex structure on the dual space $V^\vee$: this is the dual (transpose) endomorphism: $I^\vee \alpha(v)=\alpha(Iv)$. Its complexification, $I^\vee_\mathbb{C}$, induces an eigenspace decomposition of $V^\vee\otimes\mathbb{C}$.

For $\mathbb{R}^2 = (\mathbb{R}^2) ^\vee$ the "standard" complex structure $I$ is represented by the matrix $\left(\begin{array}{rr} 0&-1\\ 1&0\\ \end{array}\right)$, and $I^\vee$ by $\left(\begin{array}{rr} 0&1\\ -1&0\\ \end{array}\right)$. Notice that $I$ is skew-symmetric and the $\pm i$ eigenspaces of $I$ and $I^\vee$ are interchanged!

If you apply the above isomorphism $V\simeq V^{1,0}$ ( resp. $V\simeq V^{0,1}$) to the first standard basis vector $e_1\in\mathbb{R}^2$, you will get $$\left\{ \frac{1}{2} \left( \begin{array}{r} 1\\ -i\\ \end{array} \right), \frac{1}{2} \left(\begin{array}{r} 1\\ i\\ \end{array}\right) \right\}, $$ an eigenbasis (for $I_\mathbb{C}$) of $\mathbb{C}^2$, where the eigenvalues are ordered $\{+i,-i\}$. Its dual basis is $$ \left\{ \left(\begin{array}{r} 1\\ i\\ \end{array} \right), \left(\begin{array}{r} 1\\ -i\\ \end{array} \right) \right\}, $$ consisting of eigenvectors for $I^\vee_\mathbb{C}$, with eigenvalues $\{+i,-i\}$.

Now rephrase all of the above in terms derivations. The $\mathbb{C}$-isomorphism $V\simeq V^{1,0}$ gives you $\frac{\partial}{\partial x}\mapsto \frac{1}{2} \left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right)$, and $$\left\{ \frac{1}{2}\left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right), \frac{1}{2}\left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right) \right\}$$ is a $\mathbb{C}$-basis of $\mathbb{R}^2\otimes \mathbb{C}$, dual to $\left\{dx + i dy, dx-idy \right\}$. As the latter is conventionally denoted by $\left\{dz,d\overline{z}\right\}$, it is natural to denote its dual basis by $\left\{ \frac{\partial}{\partial z}, \frac{\partial}{\partial \overline{z}}\right\}$.

Notice that with this convention $dz\left(\frac{\partial}{\partial \overline{z}} \right) =0$ and $d\overline{z}\left(\frac{\partial}{\partial \overline{z}} \right) =1$, which is highly desirable, as mentioned in the comments.

ADDENDUM

In sum, we have to make a compatible choice of:

1) Eigenvectors of $I$
2) Eigenvectors of $I^\vee$.

We tend to put bars on the eigenvectors with eigenvalue $(-i)$. Eigenvectors are determined up to a nonzero scalar, so a priori we have to choose 4 of these (in $\mathbb{C}^2=\mathbb{R}^2\otimes\mathbb{C}$ ). However, the conditions:

a) $\frac{\partial}{\partial \overline{z}} = \overline{\frac{\partial}{\partial z}}$
b)$\left\{ \frac{\partial}{\partial z}, \frac{\partial}{\partial \overline{z}}\right\}$ is a dual $\mathbb{C}$-basis to $\left\{dz,d\overline{z}\right\}$
leave us only with the freedom of replacing $\frac{\partial}{\partial z}$ by a scalar multiple.

As an example, if you take the complex structure $\left(\begin{array}{rr} -1&-2\\ 1&1\\ \end{array}\right)$, then, up to that single scalar ambiguity, we have $$ dz = dx+ (1+i)dy, \ d\overline{z} = dx + (1-i) dy$$ $$\frac{\partial}{\partial z}=\frac{1}{2}\left( (1+i)\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right), \frac{\partial}{\partial \overline{z}}=\frac{1}{2}\left( (1-i)\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right)$$