Okay...I think this answer finally takes care of the difficulty. Thanks to arctictern for all his help :)
Let $g \in (G_{1}\times G_{2})^{\prime}$. Then, since $(G_{1} \times G_{2})^{\prime}$ is the subgroup generated by all commutators of elements of $G_{1} \times G_{2}$, $g$ is the product of commutators of elements of $G_{1} \times G_{2}$. Letting $(a_{i},b_{i})$, $(c_{i},d_{i})\in G_{1} \times G_{2}$, we have that
$\begin{align} g = [(a_{1},b_{1}),(c_{1},d_{1})]\cdot [(a_{2},b_{2}),(c_{2},d_{2})]\cdot\,\, \cdots \,\, \cdot[(a_{t},b_{t}),(c_{t},d_{t})]\,\text{for some}\, t, \\ = (a_{1},b_{1})^{-1}(c_{1},d_{1})^{-1}(a_{1},b_{1})(c_{1},d_{1})\cdot (a_{2},b_{2})^{-1}(c_{2},d_{2})^{-1}(a_{2},b_{2})(c_{2},d_{2})\cdot \, \, \cdots \, \,\cdot(a_{t},b_{t})^{-1}(c_{t},d_{t})^{-1}(a_{t},b_{t})(c_{t},d_{t}) \\ = (a_{1}^{-1},b_{1}^{-1}) (c_{1}^{-1},d_{1}^{-1})(a_{1},b_{1})(c_{1},d_{1})\cdot (a_{2}^{-1},b_{2}^{-1})(c_{2}^{-1},d_{2}^{-1})(a_{2},b_{2})(c_{2},d_{2})\cdot \, \, \cdots \, \,\cdot(a_{t}^{-1},b_{t}^{-1})(c_{t}^{-1},d_{t}^{-1})(a_{t},b_{t})(c_{t},d_{t}) \\ = (a_{1}^{-1}c_{1}^{-1}a_{1}c_{1}a_{2}^{-1}c_{2}^{-1}a_{2}c_{2}\cdots a_{t}^{-1}c_{t}^{-1}a_{t}c_{t}, \,b_{1}^{-1}d_{1}^{-1}b_{1}d_{1}b_{2}^{-1}d_{2}^{-1}b_{2}d_{2}\cdots b_{t}^{-1}d_{t}^{-1}b_{t}d_{t}) \\ = ([a_{1},c_{1}]\cdot [a_{2},c_{2}]\cdot \,\, \cdots \, \, \cdot [a_{t},c_{t}], \,[b_{1},d_{1}]\cdot [b_{2},d_{2}]\cdot \,\, \cdots \, \, \cdot [b_{t},d_{t}]) \in G_{1}^{\prime} \times G_{2}^{\prime}\end{align}$
So, we have the inclusion $\mathbf{(G_{1}\times G_{2})^{\prime}\subseteq G_{1}^{\prime}\times G_{2}^{\prime}}$
In the other direction, let $h \in G_{1}^{\prime} \times G_{2}^{\prime}$.
To belong in $ G_{1}^{\prime} \times G_{2}^{\prime}$, $h$ must be the direct product of an element of $G_{1}^{\prime}$, which is the (group operation) product of commutators of elements of $G_{1}$, and an element of $G_{2}^{\prime}$, which is the (group operation) product of commutators of elements of $G_{2}$.
Let $h_{1}$ be such an element of $G_{1}^{\prime}$ and let $h_{2}$ be such an element of $G_{2}^{\prime}$, and let $h = h_{1} \times h_{2}$. Then,
$h_{1} = [a_{1},c_{1}]\cdot [a_{2}, c_{2}] \cdot \, \, \cdots \, \, \cdot [a_{t},c_{t}]$ for some $t$, where each $a_{i}$, $c_{i}$ is an element of $G_{1}$
and $h_{2} =[b_{1},d_{1}]\cdot [b_{2}, d_{2}] \cdot \, \, \cdots \, \, \cdot [b_{t},d_{t}]$ for some $t$, where each $b_{i}$, $d_{i}$ is an element of $G_{2}$.
So,
$\begin{align}h = h_{1} \times h_{2} = ([a_{1},c_{1}]\cdot [a_{2}, c_{2}] \cdot \, \, \cdots \, \, \cdot [a_{t},c_{t}], [b_{1},d_{1}]\cdot [b_{2}, d_{2}] \cdot \, \, \cdots \, \, \cdot [b_{t},d_{t}])\\ =(a_{1}^{-1}c_{1}^{-1}a_{1}c_{1}a_{2}^{-1}c_{2}^{-1}a_{2}c_{2}\cdots a_{t}^{-1}c_{t}^{-1}a_{t}c_{t}, \,b_{1}^{-1}d_{1}^{-1}b_{1}d_{1}b_{2}^{-1}d_{2}^{-1}b_{2}d_{2}\cdots b_{t}^{-1}d_{t}^{-1}b_{t}d_{t}) \\ = (a_{1}^{-1},b_{1}^{-1}) (c_{1}^{-1},d_{1}^{-1})(a_{1},b_{1})(c_{1},d_{1})\cdot (a_{2}^{-1},b_{2}^{-1})(c_{2}^{-1},d_{2}^{-1})(a_{2},b_{2})(c_{2},d_{2})\cdot \, \, \cdots \, \,\cdot(a_{t}^{-1},b_{t}^{-1})(c_{t}^{-1},d_{t}^{-1})(a_{t},b_{t})(c_{t},d_{t}) \\ = (a_{1},b_{1})^{-1}(c_{1},d_{1})^{-1}(a_{1},b_{1})(c_{1},d_{1})\cdot (a_{2},b_{2})^{-1}(c_{2},d_{2})^{-1}(a_{2},b_{2})(c_{2},d_{2})\cdot \, \, \cdots \, \,\cdot(a_{t},b_{t})^{-1}(c_{t},d_{t})^{-1}(a_{t},b_{t})(c_{t},d_{t}) \\ = [(a_{1},b_{1}),(c_{1},d_{1})]\cdot [(a_{2},b_{2}),(c_{2},d_{2})]\cdot\,\, \cdots \,\, \cdot[(a_{t},b_{t}),(c_{t},d_{t})] \in (G_{1}\times G_{2})^{\prime} \end{align}$
So, we have the inclusion $\mathbf{G_{1}^{\prime}\times G_{2}^{\prime} \subseteq (G_{1}\times G_{2})^{\prime}}$.
Thus, since we have inclusion in both directions, we have established the equality $\mathbf{(G_{1}\times G_{2})^{\prime} = G_{1}^{\prime}\times G_{2}^{\prime}}$ for the right things this time, hopefully.
Best Answer
If $a,b\in G$ then $ab=ba$ is equivalent to $aba^{-1}b^{-1}=e$ and so the commutator of $a$ and $b$ is a measure of the failure of $a$ and $b$ commuting with each other. However, as a measure alone, it is somewhat binary: either group elements commute or they don't.
There are two ways we can quantify this measure. The first is by looking at (subgroups of) the symmetric group. Then, instead of asking if $a$ and $b$ commute, we can ask how many elements are moved around by their commutator.
The second way is to look at the commutator subgroup as a measure of how noncommutative a group is. A group is commutative if it has a trivial commutator subgroup (and highly noncommutative if the commutator subgroup is the entire group). But we can then ask if the commutator subgroup is abelian. And if not, if it's commutator subgroup is abelian. And we can repeat the process of taking commutator subgroups many times and ask if we ever find the trivial group, and if so, how many times does it take? And if not, how many times does it take before we keep on getting the same group repeating itself? This gives the notion of solvable groups and their degree of sol ability. A similar notion gives nilpotent groups and their degree of nilpotency. Both are measures of how close to abelian our groups actually are.