First question: A function is (complex) analytic if and only if it is holomorphic. It is an easy exercise to see that if $f$ is complex differentiable in $w$ and $f(w) \neq 0$, then $1/f$ is also complex differentiable in $w$ and $(1/f)'(w) = -f'(w)/f(w)^2$. So we know that $1/f$ is holomorphic in $B(z_0,\delta)\setminus \{z_0\}$, hence analytic.
Second question: It is not correct that $z_0$ is an isolated singularity of $1/f$ since $\lim\limits_{z\to z_0} \frac{1}{f(z)} = 0$. It is an isolated singularity since $1/f$ is holomorphic in $B(z_0,\delta)\setminus\{z_0\}$. The limit says that $z_0$ is a removable singularity, and the value which removes the singularity is $0$, so
$$g(z) = \begin{cases}\frac{1}{f(z)} &, z \neq z_0\\ 0 &, z = z_0 \end{cases}$$
is holomorphic on $B(z_0,\delta)$.
Third question: Since $g(z_0) = 0$, the power series representation of $g$ about $z_0$ has a constant term zero, hence
$$g(z) = \sum_{k=1}^\infty a_k (z-z_0)^k.$$
since $g$ does not vanish identically, not all coefficients are $0$, and if we let $m = \min \{ k \in\mathbb{N} : a_k \neq 0\}$, we have
$$g(z) = \sum_{k=m}^\infty a_k (z-z_0)^k$$
with $a_m \neq 0$. Then
$$h(z) = \sum_{k=m}^\infty a_k (z-z_0)^{k-m} = \sum_{r=0}^\infty a_{r+m}(z-z_0)^r$$
is holomorphic in $B(z_0,\delta)$ with $h(z_0) = a_m \neq 0$, and evidently $g(z) = (z-z_0)^m\cdot h(z)$. Since $g$ has no zero in $B(z_0,\delta)\setminus\{z_0\}$, $h$ does not vanish anywhere in $B(z_0,\delta)$, so $\tilde{h}(z) = \frac{1}{h(z)}$ is holomorphic on $B(z_0,\delta)$, and we have
$$f(z) = (z-z_0)^{-m}\cdot \tilde{h}(z)$$
on $B(z_0,\delta)\setminus\{z_0\}$.
Best Answer
A removable singularity of a function $f$ is a point $z_0$ where $f(z_0)$ is undefined, but there exists a value $c$ such that, if we define $f(z_0) = c$, then $f$ is analytic in a neighborhood of $z_0$. Note that $f$ is not actually analytic at $z_0$--it is undefined. It's just that there's a way to define its value at $z_0$ to make it analytic.
What the lemma is proving is that if a function $f$ is analytic and bounded on the set $0 < |z-z_0| < \epsilon$ for some positive $\epsilon$, then either $f$ is analytic at $z_0$, or $f$ has a removable singularity there (and thus could be made analytic through a suitable choice of $f(z_0)$). Bounded is needed because $z_0$ could otherwise be a pole or essential singularity, where no choice of $f(z_0)$ could make $f$ analytic there, but in both of those cases $f$ would be unbounded on $0 < |z-z_0| < \epsilon$.