The generating function proof of Observation 1.
Write $$f(x,z)=\sum_{n=0}^{\infty}(1+x)^nz^n=\frac{1}{1-z(x+1)}$$ Then the coefficient of $x^a$ is $$h_a(z)=\sum_{k=a}^\infty \binom ka z^k$$
When $z=1/2,$ then $h_a(1/2)$ is your value.
But $$f(x,1/2)=\frac1{1/2-x/2}=\frac{2}{1-x}$$
So $h_a(1/2)=2,$ the result you want.
For any particular value $|z_0|<1,$ you get:
$$
\begin{align}
f(x,z_0)&=\frac{1}{1-z_0}\frac1{1-x\frac{z_0}{1-z_0}}\\&=\sum_{n=0}^{\infty}\frac{z_0^n}{(1-z_0)^{n+1}}x^n
\end{align}
$$
So $$h_a(z_0)=\frac{z_0^a}{(1-z_0)^{a+1}}\tag1$$
The probability proof of Observation 1.
Let $p_n$ be the probability that in an infinite sequence of tosses of a fair coin, that after $n$ toss you have $a$ heads and the next toss is a another heads. So $$p_n=\binom na \frac1{2^{n+1}}.$$ Since it is probability zero that you will never get $a+1$ heads, this means:
$$\sum_{n} p_n=1.$$
If the probality of heads is $p,$ then:
$$p_n=\binom na p^{a+1}(1-p)^{n-a}$$
This lets you get, with a little manipulation, the same result as $(1)$ with $z_0=1-p.$
Observation 2.
If $x_n=\binom na\frac1{2^n},$ then when $n\geq a,$ we get a recurrence:
$$x_{n+1}=\frac{n+1}{2(n+1-a)}x_n.$$
Show that if $a\leq n<2a-1,$ then $2(n+1-a)<n+1,$ so $x_{n+1}>x_n.$
When $n=2a-1,$ you get $2(n+1-a)=n+1.$ So $x_{2a-1}=x_{2a}.$
When $n>2a-1$ you likewise get $x_{n+1}<x_n.$
Observation 3.
$$\sum_{k=a}^{2a} \binom ka \frac1{2^{k+1}}\tag2$$ is the probability after $2a+1$ tosses of a fair coin, that you’ve gotten $a+1$ or more heads, with $k$ being the point where the $k+1$st toss was when you got the $a+1$st heads.
But $(2)$ is also equal to the probability that you got $a+1$ or more tails.
And you always get $a+1$ or more heads or $a+1$ or more tails in $2a+1$ tosses, but not both. So (2) is equal to $\frac12.$ Multiply by $2$ and you get your identity.
Observation 4.
This is roughly the same as for observation $(3).$
Let $b=n-a.$
Then $$\sum_{k=a}^{a+b} \binom ka \frac1{2^{k+1}}\tag3$$
is the probability in $a+b+1$ tosses of a fair coin that you get $a+1$ or more heads.
And:
$$\sum_{k=b}^{a+b}\binom kb\frac1{2^{k+1}}\tag4$$
is the probability of $b+1$ or more tails in $a+b+1$ tosses of a fair coin.
As with Observation 3, exactly one of the events counted in $(3)$ and $(4)$ is true, so the sum of these two values is $1.$
Multiplying that sum by $2$ gives you the Observation 4.
If $p$ is the probability of heads in a coin, you get a generalization of Observation 4:
$$p^{a+1}\sum_{i=0}^{b}\binom {a+i}a (1-p)^{k-a}+(1-p)^{b+1}\sum_{j=0}^{a}\binom {b+j}b p^{j}=1.$$
Another way to write the probability that you get more than $a$ heads win $a+b+1$ tosses is:
$$\sum_{k=a+1}^{a+b+1}\binom{a+b+1}{k}p^k(1-p)^{a+b+1-k}\\=p^{a+1}\sum_{j=0}^{b}\binom {a+b+1}jp^{b-j}(1-p)^j$$ where $k$ goes over all the possible numbers of heads $>a,$ and $j=a+b+1-k$ loops over the possible numbers of tails.
That gives a new identity, swapping $p$ and $1-p,$ you get:
$$\sum_{j=0}^{b}\binom {a+b+1}jp^j(1-p)^{b-j}= \sum_{i=0}^{b}\binom {a+i}a p^{i}$$
When $p=\frac12,$ this gives you: $$\frac{1}{2^{a+b}}\sum_{j=0}^b \binom{a+b+1}j=\sum_{k=a}^{a+b}\binom ka\frac{1}{2^k}.\tag 5$$
Indeed, the identity $(5)$ can be seen as the heart of the all the observations other than $2.$
For example, looking at the left side, we see it is $\frac{1}{2^{a+b}}\left(2^{a+b+1}-g(b)\right),$ where $g(b)=\sum_{i=0}^a\binom{a+b+1}i$ is a polynomial of degree $a.$ So we get as $b\to \infty,$ that the right side converges to $2.$
You can prove $(5)$ directly by induction.
When $b=0,$ it is easy.
$$
\begin{align}
\sum_{j=0}^{b+1}\binom{a+b+2}j&=\sum_{j=0}^{b+1}\left(\binom{a+b+1}j+\binom{a+b+1}{j-1}\right)\\&=\binom{a+b+1}{b+1}+2\sum_{j=0}^b\binom {a+b+1}j\\&= \binom{a+b+1}{a}+2\sum_{j=0}^b\binom {a+b+1}j
\end{align}
$$ Dividing by $2^{a+b+1}$ on both sides, you get the left side for $(5)$ increases from $b$ to $b+1$ by $\frac1{2^{a+b+1}}\binom{a+b+1}{a}.$
I’m not sure the induction proof of $(5)$ is the most edifying - it is clearer if you understand the underlying probabilities. But it might be the most direct.
