[proofplan] The quotient $A_f$ is constructed precisely so that every Hecke operator in the annihilator ideal $I_f$ becomes zero on it. Therefore the Hecke action of $\mathbb{T}$ descends to an action of the quotient ring $\mathbb{T}/I_f$. The eigenvalue homomorphism identifies this quotient with the Hecke eigenvalue order generated by the coefficients of $f$, and the class of $T_n$ is exactly $a_n(f)$. [/proofplan]

[step:Show that the ideal $I_f$ acts as zero on the quotient $A_f$]Let \begin{align*} \pi:J_0(N)\to A_f=J_0(N)/(I_fJ_0(N)) \end{align*} denote the quotient morphism. By definition, $I_fJ_0(N)$ is the abelian subvariety of $J_0(N)$ generated by the images $v(J_0(N))$ as $v$ ranges over $I_f$. Hence, for every element $u\in I_f$, the image of the endomorphism \begin{align*} u:J_0(N)\to J_0(N) \end{align*} is contained in $I_fJ_0(N)$. Therefore \begin{align*} \pi\circ u=0. \end{align*} Thus every element of $I_f$ acts as the zero endomorphism on $A_f$.[/step]

[guided]The quotient $A_f$ is formed by collapsing the abelian subvariety $I_fJ_0(N)$. By definition, this subvariety is generated by all images $v(J_0(N))$ where $v\in I_f$. Therefore any point of $J_0(N)$ obtained by applying an element of $I_f$ becomes zero after passing to the quotient. Formally, let \begin{align*} \pi:J_0(N)\to A_f=J_0(N)/(I_fJ_0(N)) \end{align*} be the quotient morphism. If $u\in I_f$, then the endomorphism \begin{align*} u:J_0(N)\to J_0(N) \end{align*} has image contained in $I_fJ_0(N)$, because $I_fJ_0(N)$ is generated by all images $v(J_0(N))$ with $v\in I_f$. Therefore applying $\pi$ after $u$ kills every point: \begin{align*} \pi\circ u=0. \end{align*} So the action of $u$ on the quotient $A_f$ is the zero action. This proves that $I_f$ acts as zero on $A_f$.[/guided]

[step:Descend the Hecke action to the quotient and factor it through $\mathbb{T}/I_f$]First we verify that every Hecke operator preserves the subvariety being quotiented out. Let $t\in\mathbb{T}$. Since $\mathbb{T}$ is commutative and $I_f$ is an ideal of $\mathbb{T}$, for every $u\in I_f$ we have $tu=ut\in I_f$. Therefore \begin{align*} t(u(J_0(N)))=(tu)(J_0(N))\subseteq I_fJ_0(N). \end{align*} Because $I_fJ_0(N)$ is generated by the images $u(J_0(N))$ with $u\in I_f$, the endomorphism $t:J_0(N)\to J_0(N)$ satisfies \begin{align*} t(I_fJ_0(N))\subseteq I_fJ_0(N). \end{align*} Hence $t$ induces an endomorphism of the quotient $A_f$. This gives an action homomorphism \begin{align*} \rho:\mathbb{T}\to \operatorname{End}(A_f). \end{align*} By the previous step, every $u\in I_f$ satisfies $\rho(u)=0$, so $I_f\subseteq \ker(\rho)$. Hence there is a unique ring homomorphism \begin{align*} \overline{\rho}:\mathbb{T}/I_f\to \operatorname{End}(A_f) \end{align*} such that $\rho=\overline{\rho}\circ q$, where \begin{align*} q:\mathbb{T}\to \mathbb{T}/I_f \end{align*} is the quotient map.[/step]

[guided]Before using an action of $\mathbb{T}$ on $A_f$, we must prove that the action on $J_0(N)$ descends to the quotient. The quotient is by $I_fJ_0(N)$, so the condition to check is that every Hecke operator sends this subvariety into itself. Let $t\in\mathbb{T}$. The subvariety $I_fJ_0(N)$ is generated by the images $u(J_0(N))$ as $u$ ranges over $I_f$. Thus it is enough to check what $t$ does to each such image. If $u\in I_f$, then $\mathbb{T}$ is commutative and $I_f$ is an ideal, so $tu=ut\in I_f$. Therefore \begin{align*} t(u(J_0(N)))=(tu)(J_0(N))\subseteq I_fJ_0(N). \end{align*} Since this holds for every generator image $u(J_0(N))$, the endomorphism $t:J_0(N)\to J_0(N)$ preserves the generated abelian subvariety: \begin{align*} t(I_fJ_0(N))\subseteq I_fJ_0(N). \end{align*} This is exactly the condition that allows $t$ to induce a well-defined endomorphism of the quotient $A_f=J_0(N)/(I_fJ_0(N))$. Therefore the Hecke action on $J_0(N)$ descends to an action homomorphism \begin{align*} \rho:\mathbb{T}\to \operatorname{End}(A_f). \end{align*} The previous step showed that every element of $I_f$ acts as zero on $A_f$, so $I_f\subseteq\ker(\rho)$. By the universal property of quotient rings, there is a unique ring homomorphism \begin{align*} \overline{\rho}:\mathbb{T}/I_f\to \operatorname{End}(A_f) \end{align*} satisfying $\rho=\overline{\rho}\circ q$, where \begin{align*} q:\mathbb{T}\to \mathbb{T}/I_f \end{align*} is the quotient map.[/guided]

[step:Identify the quotient Hecke algebra with the eigenvalue order of $f$]By definition, the eigenvalue homomorphism \begin{align*} \lambda_f:\mathbb{T}\to \mathcal O_f \end{align*} sends each Hecke operator $T_n$ to its eigenvalue $a_n(f)$. Its kernel is $I_f$, so it induces an injective homomorphism \begin{align*} \overline{\lambda}_f:\mathbb{T}/I_f\to \mathcal O_f \end{align*} given by \begin{align*} \overline{\lambda}_f(t+I_f)=\lambda_f(t). \end{align*} Thus we may identify $\mathbb{T}/I_f$ with the subring $\lambda_f(\mathbb{T})\subseteq \mathcal O_f$.[/step]

[guided]The map that records the Hecke eigenvalues of $f$ is \begin{align*} \lambda_f:\mathbb{T}\to \mathcal O_f, \end{align*} defined by requiring \begin{align*} \lambda_f(T_n)=a_n(f) \end{align*} for every $n\geq 1$. The ideal $I_f$ is defined as $\ker(\lambda_f)$. Therefore the quotient map property gives a well-defined induced homomorphism \begin{align*} \overline{\lambda}_f:\mathbb{T}/I_f\to \mathcal O_f, \qquad \overline{\lambda}_f(t+I_f)=\lambda_f(t). \end{align*} This induced homomorphism is injective because its kernel consists of the classes $t+I_f$ with $\lambda_f(t)=0$, equivalently $t\in I_f$, hence only the zero class. So $\mathbb{T}/I_f$ is identified with the subring $\lambda_f(\mathbb{T})$ of $\mathcal O_f$. Under this identification, a Hecke operator is represented by its eigenvalue on $f$.[/guided]

[step:Read off the scalar by which $T_n$ acts on $A_f$] Use the injective homomorphism $\overline{\lambda}_f$ to define the induced action of the eigenvalue order $\lambda_f(\mathbb{T})\subseteq\mathcal O_f$ on $A_f$ by the map \begin{align*} \lambda_f(\mathbb{T})&\to \operatorname{End}(A_f),\\ \alpha&\mapsto \overline{\rho}(\overline{\lambda}_f^{-1}(\alpha)). \end{align*} This is well-defined because $\overline{\lambda}_f:\mathbb{T}/I_f\to\lambda_f(\mathbb{T})$ is an isomorphism onto its image. Fix $n\geq 1$. The class of $T_n$ in $\mathbb{T}/I_f$ is sent by $\overline{\lambda}_f$ to \begin{align*} \overline{\lambda}_f(T_n+I_f)=\lambda_f(T_n)=a_n(f). \end{align*} Since $\rho(T_n)=\overline{\rho}(T_n+I_f)$, the endomorphism induced by $T_n$ on $A_f$ is precisely the endomorphism associated to $a_n(f)\in \lambda_f(\mathbb{T})\subseteq \mathcal O_f$ under the displayed action. Therefore $T_n$ acts on $A_f$ through the scalar $a_n(f)$ for every $n\geq 1$, as claimed. [/step]