I was given the following proof of the statement:
Theorem: Let $X$ be a $G$-space and $Y$ be a $H$-space. Then the space $(X \times Y) / (G \times H)$ is homeomorphic to $(X / G) \times (Y / H)$
Proof: The idea is to prove that there is a homomorphism $$ \mathrm{Orb}_{G \times H} ((x,y)) \in \frac{X \times Y}{G \times H} \mapsto \mathrm{Orb}_{G}(x) \times \mathrm{Orb}_H (x) \in \frac{X}{G} \times \frac{Y}{H}.$$ In order the justify this, we consider the function $$ f: [(x,y)] \in \frac{X \times Y}{G \times H} \mapsto ([x], [y]) \in \frac{X}{G} \times \frac{Y}{H}, $$ which is well-defined since for $[(g_1, h_1) \cdot (x, y)] = [(g_2, h_2) \cdot (x, y)]$, we have \begin{align*} f([(g_1, h_1) \cdot (x, y)]) &= f([g_1 \cdot x, h_1 \cdot y]) = f([g_1 x , h_1 y])\\ &= ([g_1 x], [h_1 y]) = ([g_1 \cdot x], [h_1 \cdot y]) = ([x], [y]) \end{align*} and \begin{align*} f([(g_2, h_2) \cdot (x,y)]) &= f([(g_2 \cdot x ,h_2 \cdot y)]) = f([(g_2 x, h_2 y])\\ &= ([g_2 x], [h_2 y ]) = ([g_2 \cdot x], [h_2 \cdot y]) = ([x], [y]). \end{align*} So $f([(g_1, h_1) \cdot (x,y)]) = f([(g_2,h_2) \cdot (x,y)])$.
Similarly, we define $$ g : ([x], [y]) \in \frac{X}{G} \times \frac{Y}{H} \mapsto g([x],[y]) = [(x,y)] \in \frac{X \times Y }{G \times H}$$ and check with the same strategy that it is well-defined. Now to show that these functions are bijective inverses of one another we take $[(x,y)] \in (X \times Y)/(G \times H) $ and note that $(g \circ f)([(x,y)]) = g\left( f([(x,y)])\right) = g\left([x], [y]\right) = [(x,y)]$ and by analogy that $(f \circ g)([x],[y]) = ([x], [y])$. The assumption of having $X$ a $G$-space and $Y$ a $H$-space show that $X \times Y$ is in fact a $(G \times H)$-space so that both $f$ and $g$ are continuous and we have constructed the required homomorphism.
I am confused as to why the author insists on showing:
If $[(g_1, h_1) \cdot (x, y)]= [(g_2, h_2) \cdot (x,y)]$, then $f([(g_1, h_1) \cdot (x, y)])= f([(g_2, h_2) \cdot (x,y)])$
to show that $f$ is well-defined. Is it not sufficient to show: If $[(g_1, h_1) \cdot (x, y)] = [(x,y)]$, then $f([(g_1, h_1) \cdot (x, y)]) = f([(x,y)])$?
$f$ is defined on orbits of $G \times H$, so any two points on the orbit of $(x,y)$ under $G \times H$ must get the same image. And this is what is checked.
It's similar to when you want to define a map on classes mod $R$ to a set with classes mod $S$; you check that $xRx'$ implies $f(x)Sf(x')$ and then the definition $f([x]_R) = [f(x)]_S$ is well-defined (i.e. does not depend on the chosen representative).