Generalized Ramsey Numbers In "Graph Theory with Applications" by Bondy and Murty, the generalized Ramsey Numbers are defined as below.

Let $G_1,G_2,\dots,G_m$ be simple graphs. The generalised Ramsey Number $r(G_1,G_2,\dots,G_m)$ is the smallest integer $n$ such that every $m$-edge colouring $(E_1,E_2,\dots,E_m)$ of $K_n$ contains, for some $i$, a subgraph isomorphic to $G_i$ in colour $i$.

Say we have two simple graphs $G_1$ and $G_2$ such that $r(G_1,G_2)=5$. Then every possible $2$-edge coloring of $K_5$ has $G_1$ as a subgraph on color $1$ or $G_2$ as a subgraph in color $2$. Am I understanding this correctly?
If I am, then that would imply that for any $m$ simple graphs $G_1,G_2,\dots,G_m$, the following equality holds:
$$r(G_1,G_2,\dots,G_m)=\min\{r(G_i,G_i,\dots,G_i):i\in\{1,2,\dots,m\}\},$$ 
where there are $m$ of the $G_i$'s, correct?

To your first question, yes, your understanding of the notation is correct.
To your second question, Steve Kass has pointed out in a comment that $$r(K_3,K_2)\gt\min(r(K_3,K_3),r(K_2,K_2)).$$
This example leaves open the possibility that the inequality $$r(G_1,G_2)\ge\min(r(G_1,G_1),r(G_2,G_2))$$ holds in general. This is also false. The question is discussed on pp. 43–44 of the book Examples and Counterexamples in Graph Theory by Michael Capobianco and John C. Molluzzo. It was conjectured by Frank Harary that 

if $F_1,F_2$ have no isolates, then $r(F_1,F_2)\ge\min(r(F_1,F_1),r(F_2,F_2))$.

The counterexample to Harary's conjecture (attributed to Galvin) is that
$$r(P_5,K_{1,3})=5$$
while
$$r(P_5,P_5)=r(K_{1,3},K_{1,3})=6.$$
In order to refute Harary's conjecture, we need only show that $r(P_5,K_{1,3})\le5$ while $r(P_5,P_5)\gt5$ and $r(K_{1,3},K_{1,3})\gt5.$ (It's quite easy to see that $r(P_5,K_{1,3})\ge5$ and $r(K_{1,3},K_{1,3})\le6$ but we don't need this.)
To see that $r(P_5,P_5)\gt5,$ note that neither the graph $K_{1,4}$ nor its complement $\overline{K_{1,4}}=K_1+K_4$ contains a $P_5.$
To see that $r(K_{1,3},K_{1,3})\gt5,$ note that the self-complementary graph $C_5=\overline{C_5}$ contains no $K_{1,3}.$
To see that $r(P_5,K_{1,3})\le5,$ consider a graph $G$ of order $5$ whose complement $\overline G$ contains no $K_{1,3};$ we have to show that $G$ contains a $P_5.$ Since the graph $G$ has minimum degree $\delta(G)\ge2,$ it contains a cycle. We consider three cases.
If $G$ contains a $C_5,$ it clearly contains a $P_5.$
If $G$ contains a $C_4$ then, since the fifth vertex is joined to some vertex of the $C_4,$ we ckearly have a $P_5.$
Finally, suppose the longest cycle in $G$ is a $C_3.$ Let $u,v$ be the vertices off the $C_3.$ Then $u$ can't be joined to two vertices of the $C_3$ as that would make a $C_4;$ so $u$ is joined to $v$ and to one vertex of the $C_3,$ making a $P_5.$
