Intuitively, for a function from $f: V \rightarrow \hat{\mathbb{C}}$ to be holomorphic, several things must be true when we look at $f$ as a function from $V\cap \mathbb{C}$ to $\mathbb{C}$.
For every point $z$ in $V$ that is not $\infty$, either $f(z) \in \mathbb{C}$ and $f(z)$ is holomorphic as a $\mathbb{C} \rightarrow \mathbb{C}$ function there, or $f(z) = \infty$, implying that $f$ has a pole (in the usual sense) there.
Then if $\infty$ is in $V$, in order for $f$ to be holomorphic at $\infty$, we must have that $\lim_{|z| \rightarrow \infty} f(z)=c$ for some $c \in \mathbb{C}$, or that $\lim_{|z| \rightarrow \infty} f(z) = \infty$ (i.e., $f$ diverges in every direction).
(Again, we are looking at $f$ as a $\mathbb{C} \rightarrow \mathbb{C}$ function only and using the notions of divergence and holomorphicity we're already familiar with.)
Now we can look at your conditions again. The map $z \rightarrow \frac{1}{z}$ inverts the complex plane around the circle of radius 1. So complex numbers $z$ with very small magnitudes are mapped to numbers with very large magnitudes, and vice versa. If we look at the analytic continuation of this map on $\hat{\mathbb{C}} \rightarrow \hat{\mathbb{C}}$, then we have that $f(0) = \infty$ and $f(\infty) = 0$.
Condition (1) makes sure that $f$ is holomorphic everywhere where $z \in V$ and $z, f(z)\neq \infty$.
In condition (2), the inner function $z \rightarrow \frac{1}{z}$ maps the point $z = \infty$ to the point $z = 0$. So if $\infty$ was in $V$ and $f(\infty) \in \mathbb{C}$, $0$ will be in the domain of this function, and so $f(\frac{1}{z})$ being holomorphic on its domain would imply in particular that $f(1/z)$ is holomorphic at $z = 0$ and therefore that $f$ is (bounded and) holomorphic at $z = \infty$.
In condition (3), we are doing something similar, except that we are now inverting the range. So any points that were initially mapped to $\infty$ in $f$ are now mapped to $1/f = 0$. If $1/f$ is holomorphic on its domain (note that the domain does not include $z = \infty$!) then in particular, it must be holomorphic at the points $\{z \in \mathbb{C} | f(z) = \infty\}$. This fills in the gaps left by condition (1).
Condition (4) takes care of the very last case, which is the case in which $z = \infty$ is mapped to $\infty$.
I think this is a somewhat un-transparent way of defining holomorphicity on $\hat{\mathbb{C}}$. I will try to illuminate some of the intuition. We can look at $\hat{\mathbb{C}}$ as the Riemann sphere. Then consider a function $f$ that is holomorphic as a map from $V \in \hat{\mathbb{C}} \rightarrow \hat{\mathbb{C}}$. At any point on $V$, it must map to some point on $\hat{\mathbb{C}}$, it must be continuous at that point, and its derivative must exist and be continuous.
For a point $z\neq \infty$, this is equivalent to being meromorphic at $z$. For $z = \infty$, this is equivalent to the function either diverging in all directions ($f(\infty) = \infty$) or having a specific limit in all directions ($f(\infty)\in \mathbb{C}$). So a function $f: V \rightarrow \hat{\mathbb{C}}$ that is holomorphic is (a) meromorphic on the complex plane and (b) has a limit as $|z|$ goes to infinity (or diverges in all directions).
Interestingly, some functions that are entire on $\mathbb{C}$ fail to be entire here (notably $e^z$, which has an essential singularity at infinity) and some functions that are merely meromorphic on $\mathbb{C}$ (every rational function) are now entire as functions from $\hat{\mathbb{C}} \rightarrow \hat{\mathbb{C}}$. The map $z \rightarrow \frac{1}{z}$ also has a special significance on the Riemann sphere- it is equivalent to reflecting the top and bottom hemispheres.
