I'm a little confused on what Proposition 2.2 of Massey's Algebraic Topology: An Introduction is saying; I first rewrote the theorems below and then after my issues are summarized:

Theorem 2.1: If $\{G_i\}_{i\in I}$ is a collection of abelian groups, then for any abelian group $A$ and any collection of homomorphisms $\psi_i:G_i \to A$ there exists a unique homomorphism $f:G\to A$ such that for any $i\in I$ the following diagram is commutative

$$\varphi_i : G_i \to G$$ $$\psi_i:G_i\to A$$ $$f:G\to A$$ (I don't know how to draw this diagram... help if you don't mind)

Proposition 2.2: Let $\{G_i\},G$ and $\varphi_i:G_i \to G$ be as in Theorem 2.1; let $G'$ be any abelian group and let $\varphi_i':G_i \to G'$ be any collection of homomorphisms such that the conclusion of Theorem 2.1 holds with $G'$ and $\varphi_i'$ substituded for $G$ and $\varphi_i$ respectively. Then, there exists a unique isomorphism $h:G\to G'$ such that the following diagram is commutative for any $i\in I$

$$\varphi_i : G_i \to G$$ $$\varphi'_i:G_i\to G'$$ $$h:G\to G'$$

1) My first issue is that Prop 2.2 says let $\varphi_i$ be any collection of homomorphisms such that the conclusion of Theorem 2.1 holds. Isn't that any homomorphism from $G_i$ to $G'$... that was the point of Theorem 2.1, right? We get the existence of a unique $f:G\to A$ that makes the diagram commute regardless of what $\varphi_i$ are.

2) I'm confused by the "substituded for $G$ and $\varphi_i$" part. Maybe this is a matter of just having the theorem rephrased, it isn't clicking.... maybe item 1 is me not understanding the "substituded for $G$ and $\varphi_i$" part correctly?

As for the commutative diagram, I'm having a hard time typing it here on stackexchange, but it would look like this: $$\begin{array}{ccc} G_i & \xrightarrow{\phi_i} & G \\ &\searrow{\psi_i} & \downarrow{f}\\ & & A \end{array}$$
To answer your questions, another way to view the proposition that may be more intuitive is the following: If another abelian group $G'$ and an associated family of group homomorphisms $\phi_i'$ satisfy all the same properties as $G$ and $\phi_i$ in the theorem, then $G$ and $G'$ are isomorphic. That is, the properties in the theorem specify the abelian group $G$ (which is called a coproduct in the category of abelian groups) uniquely up to isomorphism.