Let us start with a generic function $f_n$ on a certain domain $S$. If we know that $||f_n||_{\infty}<C$, where $C$ is a bound such that $C<\infty$, could we state that
$$||f_n||_{\infty}<C\implies \sum_{n=0}^{\infty}f_n \text{ uniformly convergent on }S\tag{1}$$
?
If so, why? If not, could you please give/hint a counterexample?
No it does not, on the contrary there are many counterexamples - even amongst the series of numbers.
Just take $S=\{0\}$ and $f_n(0)=\frac 1n$. Certainly, $f_n$ is bounded, but the series $\sum f_n$ does not converge.
If you want to have a convergence criterium, what you are looking for is so-called normal convergence, i.e. $ \sum_n \|f_n\|_\infty$ is a convergent series. This in fact yields uniform convergence of $\sum_n f_n$. The other direction does not hold (uniform convergence of $\sum_n f_n$ does not imply convergence of $\sum_n\|f_n\|_\infty$).
Edit: I am adding your example from the comments for clarity. Therefore, let $ a_n = \sqrt{n} 2^{-n/2} $.
First, observe that $\sum a_n$ converges. This can be verified using your favorite convergence criterion for series, i.e., ratio test: $a_{n+1}/a_n=\sqrt{\frac{n+1}{n}}2^{-1/2}\to\frac1{\sqrt2}<1$.
If you now have a sequence of functions $(f_n)$ satisfying $\|f_n\|_\infty\le Ca_n$ for some fixed $C>0$ and all (except for finitely many) $n\in\mathbb N$, then $\sum f_n$ converges uniformly (because it converges normally, c.f. also the Weierstrass $M$-test).