= Solution
If $m\in T_1$, then $rm=0$ for some nonzero $r$, and
$$
r\phi(m)=\phi(rm)=0.
$$
Thus $\phi(T_1)\subseteq T_2$. The formula
$$
\boxed{\Phi(m+T_1)=\phi(m)+T_2}
$$
is consequently independent of the representative and defines an $R$-module homomorphism on the <quotient module>.
If $\phi$ is injective and $\Phi(m+T_1)=T_2$, then $\phi(m)$ is torsion, so $r\phi(m)=0$ for some nonzero $r$. Injectivity gives $rm=0$, hence $m\in T_1$. Therefore
$$
\boxed{\phi\text{ injective }\Longrightarrow\Phi\text{ injective}}.
$$
The converse fails. For $R=\mathbb Z$, take
$$
M_1=\mathbb Z\oplus\mathbb Z/2\mathbb Z,\qquad
M_2=\mathbb Z,
\qquad
\phi(n,\bar a)=n.
$$
Then $\phi$ is not injective, but quotienting $M_1$ by its torsion submodule leaves $\mathbb Z$, and $\Phi:\mathbb Z\to\mathbb Z$ is the identity.
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