I have to show this proposition:
Let $U$ and $W$ be vector spaces and $V=U \oplus W $. Then $V$ has subspaces $U'$ and $W'$ such that we can identify $U'$ with $U$, $W'$ with $W$ and such that $V$ is the internal direct sum if $U'$ and $W'$.
If I define $U'=\{(u,0) | u \in U\}$ and $ W' = \{(0,w) | w \in W \}$ and claim that these satisfy the things claimed in the proposition.
How can I identify $U'$ with $U$ by the map $U' \rightarrow U$ given by $(u,0) \mapsto u$?
It's difficult to understand exactly what you're asking, but it seems like you question is something like "what does the question mean when it says identify $U'$ with $U$?". I suspect that the answer to this question is that "identifying $U'$ with $U$" simply means showing that the two vector spaces are isomorphic, "in a natural way".
So, by defining the map $\phi:U' \to U$ by $\phi(u,0) = u$, you have given an example of a "natural" isomorphism between the spaces $U'$ and $U$. Thus, you have shown that $U'$ and $U$ are naturally isomorphic. That is, you have "identified" $U'$ with $U$.
The notion of "natural isomorphism" is made rigorous with category, but for the purposes of the question asked the fact that $\phi$ is a "sufficiently nice" map (admittedly a subjective criterion) is enough to say that we have indeed identified the two spaces as being "the same".