By definition $\binom{x}{k} = \frac{x \cdot (x - 1) \cdot \dots \cdot (x - k + 1)}{k!}$, in particular note that there are $k$ terms on the top, this is where we will get our $(-1)^k$ from if $x$ is negative. So letting $x = -\frac{1}{2}$ we get: $$(-1)^k \cdot \frac{\frac{1}{2} \cdot \frac{3}{2} \cdot \frac{5}{2} \cdot... \cdot \frac{2k - 1}{2}}{k!} = \left(\frac{-1}{2}\right)^k \frac{1 \cdot 3 \cdot... \cdot (2k - 1)}{k!}$$

Note that $2 \cdot 4 \cdot ... \cdot 2k = \prod_{i = 1}^k 2i = 2^k \cdot k!$, thus we can multiply the top and bottom of this fraction by $2^k \cdot k!$ and we get:

$$ \left(\frac{-1}{2}\right)^k \frac{1 \cdot 3 \cdot... \cdot (2k - 1)}{k!} = \left(\frac{-1}{4}\right)^k \frac{(2k)!}{(k!)^2} = \left(\frac{-1}{4}\right)^k \cdot \binom{2k}{k}$$ as desired.

For the second part, by Newton's generalized binomial theorem: $$(1 + x)^y = \sum_{k = 0}^\infty \binom{y}{k} x^k. $$

Substituting $y = -\frac{1}{2}$ and using the above formula we see that:

$$\frac{1}{(1 + x)^{\frac{1}{2}}} = \sum_{k = 0}^\infty \binom{2k}{k} \left(\frac{-x}{4} \right)^k$$

which is the formula desired.