Left adjoint to inclusion functor of torsion-free abelian groups in abelian groups

Yes, the left adjoint mods out the torsion subgroup. If $A$ is any abelian group and $B$ is torsion-free, then certainly any map $A\to B$ factors uniquely through this quotient, which is all that needs to be shown for a left adjoint to a fully faithful functor.

For every abelian group $A$ we can consider its set of torsion-elements, $$ T(A) := \{ a \in A \mid \text{there exists an nonzero $n \in \mathbb{Z}$ with $na = 0$} \} \,. $$ This is a subgroup of $A$ because $\mathbb{Z}$ is an integral domain.

The quotient $$ F(A) := A / T(A) $$ is torsion-free. Indeed, if $[a]$ is an torsion element of $F(A)$ then there exists some nonzero scalar $n \in \mathbb{Z}$ with $$ 0 = n [a] = [n a] \,. $$ This means that $na \in T(A)$, i.e. that $na$ is a torsion element of $A$. There hence exist some nonzero scalar $m \in \mathbb{Z}$ with $mna = 0$. The scalar $mn$ is again nonzero because $\mathbb{Z}$ is an integral domain. We thus find that $a$ is a torsion element of $A$, i.e. an element of $T(A)$. This shows that $[a] = 0$.

For any two abelian groups $A$ and $B$ and every homomorphism of groups $f$ from $A$ to $B$ we have $$ f(T(A)) \subseteq T(B) \,. $$ It follows that the homomorphism $f$ descends to a homomorphism $$ F(f) \colon F(A) \to F(B) \,, \quad [a]    \mapsto [f(a)] \,. $$ This construction gives a functor $$ F \colon \mathbf{Ab} \to \mathbf{TFAb} \,. $$

Suppose now that $A$ is any abelian group and that $B$ is a torsion free abelian group. If $f$ is any homomorphism from $A$ to $B$ then $$ f(T(A)) \subseteq T(B) = 0 \,, $$ so $f$ factors through a homomorphism $A / T(A) \to B$. This shows that every homomorphism from $A$ to $B$ comes from a homomorphism from $F(A)$ to $B$. In other words, the map \begin{align*} \{ \text{homomorphisms $F(A) \to B$} \} &\longrightarrow \{ \text{homomorphisms $A \to B$} \} \,,    \\ f &\longmapsto f \circ \eta_A \end{align*} is a bijection, where $\eta_A \colon A \to F(A)$ is the canonical projection given by $a \mapsto [a]$. If we denote the inclusion functor from $\mathbf{TFAb}$ to $\mathbf{Ab}$ by $R$ then we this gives us a bijection \begin{align*} \varphi_{A,B} \colon \operatorname{Hom}_{\mathrm{TFAb}}(F(A), B) &\longrightarrow \operatorname{Hom}_{\mathrm{Ab}}(A, R(B)) \,,    \\ f &\longmapsto f \circ \eta_A \,. \end{align*} This bijection is natural in both $A$ and $B$ and does therefore show that $F$ is left adjoint to $R$.