[guided]The representation produced directly by cohomology is usually denoted here by $r_{f,\lambda}$. The theorem statement asks for a semisimple representation, so we define
\begin{align*}
\rho_{f,\lambda}:G_{\mathbb Q}&\to GL_2(K_{f,\lambda})
\end{align*}
to be the semisimplification of $r_{f,\lambda}$.
This operation has three properties needed here. First, $\rho_{f,\lambda}$ is semisimple by construction. Second, trace and determinant are unchanged by semisimplification: for each $g\in G_{\mathbb Q}$, the characteristic polynomial of the action on a finite-dimensional representation equals the product of the characteristic polynomials on its Jordan-Hölder constituents, and the semisimplification has exactly the same constituents. Hence
\begin{align*}
\operatorname{tr}(\rho_{f,\lambda}(g))&=\operatorname{tr}(r_{f,\lambda}(g)),\\
\det(\rho_{f,\lambda}(g))&=\det(r_{f,\lambda}(g)).
\end{align*}
Third, unramifiedness is preserved: if $p\nmid N\ell$, inertia at $p$ acts as the identity on $r_{f,\lambda}$, so it acts as the identity on every subquotient and therefore on the semisimplification.
Applying these facts to $g=\operatorname{Frob}_p$ and using the formulas already obtained for $r_{f,\lambda}$ gives
\begin{align*}
\operatorname{tr}(\rho_{f,\lambda}(\operatorname{Frob}_p))&=a_p(f),\\
\det(\rho_{f,\lambda}(\operatorname{Frob}_p))&=\varepsilon(p)p^{k-1}.
\end{align*}
This is exactly the existence, continuity, semisimplicity, unramifiedness, trace formula, and determinant formula required in the statement.[/guided]