So I think I understand what binomial coefficients are, "K number of combinations that can be made from set N". However, I keep seeing examples demonstrating K being the number of successes against the number of outcomes.
However, my thinking is by the logic of the first example I gave, when finding the number of combinations Tails Tails would be included as it is a combination that can be made from set N. However, this does not feature a heads and is not a success.
The binomial distribution and binomial coefficients are linked, but binomial coefficients are more general.
The binomial coefficient $\binom{n}{k}$ is the number of ways to select $k$ objects from $n$ unique objects, (the number of unordered subsets with $k$ elements out of $n$ unique elements.) You can also look as $\binom{n}{k}$ as the coefficients in front of $x^k$ in the binomial expansion of $(1+x)^n$.
Let $X_1$ be a Bernoulli$(p)$ distribution: $$\mathbb{P}(X_1 = 1) = p, \quad \mathbb{P}(X_1 = 0) = 1-p.$$ You can think of a Bernoulli distribution as the result of a coin flip: 1 is a heads and comes up with probability $p$, tails comes up with probability $1-p$.
A binomial$(n,p)$ distribution $Y$ is the sum of $n$ independent Bernoulli$(p)$ distributions: $$Y = X_1 + X_2 + ... + X_n.$$
Let's look at the probability that $Y = k$ for $0 \leq k \leq n$ (there are $n$ coins flipped, and $k$ of them are heads.) First, we need to find the number of ways that $k$ coins were heads out of $n$ coins: there are $\binom{n}{k}$ ways to do so. Next, look at a specific arrangement of coins. For concreteness, suppose the first $k$ flips are heads and the $n-k$ flips after are tails, i.e., we look at the following event $$X_1= X_2 = ... = X_k = 1, X_{k+1} = X_{k+2} = ... = X_n = 0.$$ This would correspond to the subset $\{1,2,...,k\}$ of the set $\{1,2,...,n\}.$
Now, we compute the probability that the above event occurred: since each of the $X_i$ are independent Bernoulli(p) distributions, this probability is $p^k(1-p)^{n-k}$
Then, we multiply: $$\mathbb{P}(Y = k) = \binom{n}{k}p^k(1-p)^{n-k}.$$
So, we can see that the binomial coefficients are very related to the binomial distribution, but they aren't exactly the same.
If we set $p = 1-p= 1/2$, we can view the binomial distribution in a slightly more nuanced way $\mathbb{P}(Y = k)$ is $\binom{n}{k}$ divided by the total number of combinations (subsets) of $\{1,2,...,n\} = 2^n:$ it is basically saying if we pick a subset of $\{1,2,...,n\}$ at random, what is the probability that said subset is $k$ elements. However, our previous interpretation of using coins still applies.
I'm not sure what you mean by your bottom statement, suppose $n = 2$, and we flip Tails Tails. There are no successes (no heads), so the subset this corresponds to is the empty set $\{\}$, (sometimes written as $\emptyset$.) In binomial coefficient terms, this is $\binom{2}{0} = 1$.