The map $f(u+v) = u - v$ is an isomorphism on a direct product

The annoying thing with direct sums is that there are two ways of introducing it: internal and external (which is what's being mentioned in the comments). In your case, we're talking about an internal direct sum.

So, the fact that $U,W$ are vector subspaces of $V$ such that $U\cap W=\{0\}$ and $V=U+W$ means that the mapping $\alpha:U\times W\to V$, $\alpha(u,w):= u+w$ is a linear isomorphism ($U\cap W=\{0\}$ gives injectivity, and $V=U+W$ gives surjectivity by definition). This isomorphism identifies $V=U\oplus_{\text{internal}}W$ with $U\times W=U\oplus_{\text{external}}W$.

Now, consider the mapping $\phi:U\times W\to U\times W$ defined as $\phi(u,w):=(u,-w)$. You say that you're confused by the sum argument notation. Well, the definition of $f:V\to V$ is \begin{align} f:= \alpha\circ \phi\circ \alpha^{-1} \end{align}

In other words, for each $v\in V$, there exist unique $u\in U,w\in W$ such that $v=u+w$. So, $\alpha^{-1}(v)=(u,w)$, so performing the composition, we get \begin{align} f(v)&=(\alpha\circ \phi\circ \alpha^{-1})(v)=\alpha(\phi(u,w))=\alpha(u,-w)=u+(-w)=u-w \end{align} Hence, rather than explicitly introducing the notation for the isomorphism $\alpha$, your problem simply states the definition of $f$ as $f(u+w):=u-w$ where $u\in U,w\in W$.

As for proving that $f$ is indeed an isomorphism though, the more "abstract" definition $f=\alpha\circ \phi\circ \alpha^{-1}$ makes it almost obvious (why?)