Let $i\in 1,\dots n$.
Let $f_i\geq0$ be real numbers such that $\sum_{i=1}^nf_i=1$. Then what is the maximum of $B\left(\omega_i\right)=\sum_{i=1}^nf_i\omega_i$ over $\omega_i$ such that $\omega_i\geq0$ and $\sum_{i=1}^n\omega_i=1$? And which values of $\omega_i$ is it achieved at?
Initially I thought that the maximum will be when $\omega_i=f_i$ (the logic being that we think of the $f_i$ and $\omega_i$ as the component of $n$-dimensional vectors $f,\omega$, and the dot product will be maximised when they are parallel). However, in the case all the $f_i=\frac1n$, we have $B\left(\omega_i\right)=\frac1n$, so is constant. So my initial guess was wrong.
$\let\le\leqslant$Convention. Below, all $\omega = (\omega_1, \ldots, \omega_n)$ satisfy the constraints given in the problem.
Claim. For all $\omega$, we have $$B(\omega) \leqslant \max_{1 \le i \le n} f_i.$$
Proof. WLOG, $f_1 = \max_i f_i$. Then, we have \begin{align} B(\omega) &= \sum_{i = 1}^{n} f_i\omega_i \\ &= f_1(1 - \omega_2 - \cdots - \omega_n) + f_2 \omega_2 + \cdots + f_n \omega_n \\ &= f_1 + \omega_2(f_2 - f_1) + \cdots + \omega_n (f_n - f_1) \\ &\le f_1 + 0 + \cdots + 0 = f_1. &\Box \end{align}
Moreover, this equality is actually attained when $\omega_1 = 1$ and $\omega_i = 0$ for $i \geqslant 2$. (In the general case, you put $\omega_j = 1$ where $j$ is the index achieving the $\max$ for $f$ and put $\omega_i = 0$ for the remaining $i$.)
Note that analysing the above proof will show you the following: Suppose you have $$f_1 = f_2 = \cdots = f_k > f_{k + 1} \geqslant \cdots \geqslant f_n.$$ Then, you must put $\omega_i = 0$ for $i > k$ and the first $k$ $\omega$s can be anything that satisfy the property that $\sum_i \omega_i = 1$.
Here's why your initial intuition doesn't work: You are wishing to maximise $\langle f, \omega\rangle_2$ against the constraint that $\|\omega\|_1 = 1$. The norm in the constraint is different from the norm induced by the inner product. Your guess would have been correct (up to scaling) if the constraint on $\omega$ was $\sum_{i = 1}^{n} \omega_i^2 = 1$.