Center/Commutator Subgroup of Direct Product = Direct Product of these Subgroups - - Fraleigh p. 64 Theorem 6.14

(1.) What's the intuition?

Full proof for Center Subgroups

(2.) What's the proof blueprint? I know proof's using $A = B \iff A \subseteq B \wedge B \subseteq A$.
But where did $(ga,hb)$ in $(\subseteq)$ loom from? $(gh, ab)$ in $(\subseteq)$ from?

(3.) p. 2

I know commutator subgroup of $G_1 \times G_2 = \{ (g_1,g_2)(h_1,h_2), (g_1,g_2)^{-1},(h_1,h_2)^{-1} : g_i,h_i \in G_i, \forall \; i = 1,2 \; \}$
and commutator subgroup of $G_i = \{ g_ih_ig_i^{-1}h_i^{-1} : g_i,h_i \in G_i \}$.
But what's $[(g_1,g_2)(h_1,h_2)], [(g_1,h_1),e_{G_1}]$ ? Why use this perplexing notation?

Suppose $G, \ast$ and $H, \star$ are groups. Then the direct product is a group with operation $\diamond$ defined as follows: $$\diamond: (G \times H) \times (G\times H) \to (G \times H): ((g_1, h_1), (g_2, h_2)) \mapsto (g_1, h_1) \diamond (g_2, h_2) = (g_1 \ast g_2, h_1 \star h_2)$$ (you can easily check that this is indeed a group under this operation). In words, the operation on the direct product is applying the operation at each component. However, in order to shorten the notation, it is convenient to denote $g_1 \ast g_2$ by $g_1g_2$ and analogously $h_1 \star h_2$ becomes $h_1h_2$. If we now use this, it is easy to see that $(g,h)(a,b) = (ga,hb)$.
For your question about the blueprint of this prove: you mentioned it yourself: to prove an equality of sets, prove two inclusions. An inclusion $A \subset B$ is proven if you can show that each element of $A$ is in $B$. Therefore, take an arbitrary element $a \in A$ and show that $a \in B$. Note that the center of $G$, denoted by $Z(G)$ is the set of elements which commute with all elements of $G$, denoted as a set, we find: $Z(G) = \{g \in G \vert gg' = g'g \: \: \forall g' \in G\}$.
For your second question about the commutator subgroup: the commutator of two elements $x,y$ is denoted by $[x,y]$ and is defined as you stated: $[x,y] = xyx^{-1}y^{-1}$.