Count Sketch probability bound
$\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$
Denote the support of $f$ by $f_1\le f_2\le...\le f_s$. Given a frequency sequence $f$, denote by $E_{f,a}$ the random variable $\hat{f}_a$ where the stream frequencies are given by $f$ (the evaluation depends on the inner randomness of the sketch and the input stream).
The probability that the hash of one of the top $\ell=1/\epsilon^2$ elements collides with the hash of some given element $a$ is bounded by $\frac{\ell}{k}$ (union bound). Conditioning on the event that such a collision did not occur, the evaluation $\hat{f}_a$ does not depend on the top $\ell$ entries, as they do not affect the hash entry of $a$. More formally, conditioned on the event $C=\bigcap\limits_{i=0}^{\ell-1}\mathbb{1}_{h\left(f_{s-i}\right)\neq h(a)}$, $E_{f,a}$ and $E_{f^{res(\ell)},a}$ are equally distributed. This yields:
$$\Pr\left[\left|\hat{f}_a-f_a\right|\ge \epsilon\norm{f_{-a}^{res(\ell)}}\right]=
\Pr[C]\cdot \Pr\left[\left|\hat{f}_a-f_a\right|\ge \epsilon\norm{f_{-a}^{res(\ell)}}\big| C\right]+\\
\left(1-\Pr[C]\right)\Pr\left[\left|\hat{f}_a-f_a\right|\ge \epsilon\norm{f_{-a}^{res(\ell)}} \big|\overline{C}\right]\le\\
\Pr\left[\left|\hat{f}_a-f_a\right|\ge \epsilon\norm{f_{-a}^{res(\ell)}}\big| C\right]+\frac{1}{6}$$
Since $E_{f,a}, E_{f^{res(\ell)},a}$ are equally distributed (conditioned on $C$), the left hand term equals $\Pr\left[\left|\hat{f}_a-f_a\right|\ge \epsilon\norm{f_{-a}}\right]$, which you can bound by $1/6$ with your original result.