In "Categories for the Working Mathematician" by Saunders Mac Lane, chapter IV.3, p.93.
In any category $C$ a skeleton of $C$ is any full subcategory $A$ such that each object of $C$ is isomorphic (in $C$) to exactly one object of $A$. Then $A$ is equivalent to $C$ and the inclusion $K: A \to C$ is an equivalence of categories. For, select to each $c \in C$ an isomorphism $\theta_c:c\cong Tc$ with $Tc$ an object of $A$. Then we can make T a functor $T: C\to A$ in exactly one way so that $\theta$ will become a natural isomorphism $\theta : I \cong KT$.
I suppose there is an abuse of notation, and $\theta_c:c\cong KTc$.
For a morphism $f:c \to b \in C$, how do you define $Tf \in A$ ?
To verify naturality, the relation $KTf \circ \theta_c = \theta_b \circ f$ must be verified. And, since $\theta$ is a natural isomorphism, one can write $KTf = \theta_b \circ f \circ \theta_c^{-1}$. Since the functor $K$ is full, and $\theta_b \circ f \circ \theta_c^{-1}$ is a morphism $KTc \to KTb$, then there is unique preimage to $KTf$ in $A$, and that is how you define $Tf$ in $A$.
Is my argument correct ? If not, how is $Tf$ defined ?
Yes, it is correct.
About the abuse of notation, $A$ was considered to be fully contained by $C$, so $K$ is specifically the identical embedding