Why do we use dG < 0 to describe a spontaneous process? With something like a reaction or phase change, all sources use the criterion that $dG &lt; 0$ for the reaction to be spontaneous and then substitute an appropriate expression for $dG$ to the specific application. For phase change my book stats $(\mu'-\mu'')dn' &lt; 0$ where $'$ is phase A and $''$ is phase B, $\mu$ is the chemical potential and $dn' &gt; 0$ means gaining of molecules in phase A. 

However, reactions require activation energy and phase change require nucleation (I think this is kind of like an activation energy); both of these processes require an increase in $G$ before a larger decrease. Why do we not use $\Delta G$ (between initial and final equilibrium states) instead?

Edit:

I am confused by the use of dG (a rate differential) as opposed to $\Delta G$ (a net change between end states) because I'm not sure how one would integrate this for say, a phase change, since it must pass through a potential barrier where G must increase. Is the integral just $G_2-G_1$ between the end states and thermodynamically the process ignores the barrier (since thermodynamics deals with equilibrium only)? A typical thermodynamic process I'm imagining is quasistatic compression, where we can get work from the integral of PdV and P is defined as it goes through an infinite number of equilibrium states; I'm not sure if it's possible to draw a parallel here for free energy and phase change/reacting systems.   

Edit 2: 

From this graph it looks like G increases then decreases to stable equilibrium after phase change. If we find dG/dr from this graph and integrate dG(r), should we get the same $\Delta G$ as from thermodynamics (before and after phase equilibrium)? If so, this seems to me like finding work of a rapid piston-cylinder compression of an insulated system; PdV is undefined since we are not passing through a set of equilibrium states but we can still find work from $\Delta U$. Difference is that in this case, our initial state is technically not in equilibrium/stable since the system favours a phase change, so we are beginning from a state that technically does not exist on the phase diagram (like if we have a supercooled/superheated liquid).
Maybe its best to see how we generally derive the Gibbs energy... If you think about this then the answer should be clear ;) 

The Gibbs potential is a function of $P$, $T$ and $N$. That is to say:
\begin{equation}
G=G(T,P,N)
\end{equation}
Where $T$ is the temperature, $P$ is the pressure and $N$ is the number of particles.
\begin{equation}
dG=\frac{\partial G}{\partial P}\bigg|_{T,N}dP+\frac{\partial P}{\partial T}\bigg|_{P,N}dT+\sum _{i=1}^{I}\frac{\partial G}{\partial N_i}\bigg|_{P,T,N_j\neq i}dN_i
\end{equation}
Use some Maxwell relations (Look these up):
\begin{equation}
dG=VdP-SdT+\sum ^{I}_{i=1}\mu _i dN_i
\end{equation}
By definition we state the following formula at constant temperature and pressure:
\begin{equation}
dG=\sum _{i=1}^{I}\mu _idN_i
\end{equation}
Therefore you are actually just expressing the Gibbs energy in a different way when your book quotes the chemical potential. In fact the very definition of the chemical potential is:

\begin{equation}
\mu _i=\bigg(\frac{\partial G}{\partial N_i}\bigg)_{P,T,N_{j\neq i}}
\end{equation}

I don't know your mathematical ability but if it helps...its pretty clear that this is just a simple Legendre transform of the internal energy to switch between $U$ and $G$. 
(As an aside), with a little product ruling we get:
\begin{equation}
dG=\sum _{i=1}^{I}\big(\mu _idN_i+N_id\mu _i\big)
\end{equation}
This leaves us with the Gibbs-Duhem relation:
\begin{equation}
\sum ^{I}_{i=1}N_id\mu_i=-SdT+VdP
\end{equation} 
Which is a handy result. 

Also just to add on the end here about nucleation points... lets imagine a precipitation in solution....As the size of the nucleus grows (as more bits "clump" together) the Gibbs energy of the particle will rise ... this is due to two competing effects:

1) the interface interaction (increasing energy) 

2) the volume free energy (decreasing energy)

The total energy is just the resultant of these processes ... at small size the surface interactions outcompete the volume energy liberation .... as the radius of the particle increases we see a switch.