The first part of the Quantum Phase Estimation (QPE) is the implementation of this circuit (pg. 222 of Mike and Ike)
(There should be $+$ between $|0\rangle$ and $e^{2 \pi i 2^{n} }|1\rangle$ )
This gives you the state the first equation in your question:
$$
\frac{1}{2^{t/2}}(|0\rangle + e^{2\pi i 2^{t-1} \phi}|1\rangle)(|0\rangle + e^{2\pi i 2^{t-2} \phi}|1\rangle)...(|0\rangle + e^{2\pi i 2^{0} \phi}|1\rangle)
$$
To see how this is equivalent to the second question, let's suppose that $\phi$ can be written exactly as $t$ bits, and so
$$\phi = 0.\phi_1 \phi_2 \cdots \phi_t = \dfrac{\phi_1}{2^1} + \dfrac{\phi_2}{2^2} + \cdots + \dfrac{\phi_t}{2^t} = \sum_{i=1}^{t} \dfrac{\phi_i}{2^i} $$
Now note the following:
\begin{align}
2^k \phi &= \phi_1 \phi_2 \cdots \phi_k.\phi_{k+1} \phi_{k+2} \cdots \phi_{t} \\
&=\phi_1 \phi_2 \cdots \phi_k + 0.\phi_{k+1} \phi_{k+2} \cdots \phi_{t}
\end{align}
And therefore
\begin{align}
e^{2\pi i 2^k \phi} &= e^{2\pi i (\phi_1 \phi_2 \cdots \phi_k + 0.\phi_{k+1} \phi_{k+2} \cdots \phi_{t} )} \\
&=e^{2\pi i \phi_1 \phi_2 \cdots \phi_k} e^{2\pi i 0.\phi_{k+1} \phi_{k+2} \cdots \phi_{t} }\\
&= 1 \cdot e^{2\pi i 0.\phi_{k+1} \phi_{k+2} \cdots \phi_{t} } \hspace{0.75 cm} \textrm{since} \ \ e^{2\pi i \cdot n} = 1\ \ \forall n \in \mathbb{N}
\end{align}
Thus that prompts the result:
$$
\frac{1}{2^{t/2}}(|0\rangle + e^{2\pi i 0.\phi_t}|1\rangle)(|0\rangle + e^{2\pi i 0.\phi_t\phi_{t-1}}|1\rangle)...(|0\rangle + e^{2\pi i 0.\phi_t\phi_{t-1}...\phi_0}|1\rangle)
$$
that you see (formula 5.21 in Mike and Ike). Now, in term of QFT part, remember what the QFT transformation does, which is the following:
$$QFT: |\phi_n\rangle \otimes |\phi_{n-1} \otimes \rangle \otimes \cdots \otimes |\phi_1\rangle \rightarrow \dfrac{|0\rangle + e^{2\pi i 0.\phi_n}|1\rangle}{\sqrt{2}} \otimes \dfrac{|0\rangle + e^{2\pi i 0.\phi_{n-1}\phi_n}|1\rangle}{\sqrt{2}} \otimes \cdots \otimes \dfrac{|0\rangle + e^{2\pi i 0.\phi_1 \cdots \phi_{n-1} \phi_n}|1\rangle}{\sqrt{2}}$$
Which again is the state that you have at the end of the circuit above. Thus, by performing the inverse of QFT, we get back $|\phi_n\rangle \otimes |\phi_{n-1} \otimes \rangle \otimes \cdots \otimes |\phi_1\rangle$. Therefore, upon measurement on each of the qubit in the First Register, you will be able to exact $\phi_n, \cdots, \phi_1$ which allow you to reconstruct the phase $\phi$.
Hopefully I was able to help and provide some clarifications.
