That the first condition implies the first is immediate, since (using your notation) you always have $m_i \le f(\xi_i) \le M_i$, so the sums in the second definition are caught between the $L$ und $U$ sums.
Edit in response to a comment an additional explanation is necessary here. For this direction
it suffices to show that $I^* = lim_{||P||\rightarrow 0} L(f,P)$ and $I_* = lim_{||P||\rightarrow 0} U(f,P)$ Since both parts are similar it suffices to show, e.g., the first equality.
First it is easy to see that for partitions $P\subset P^\prime$ we have $L(f,P)\le L(f,P^\prime)$. A remaining hurdle is that for two partitions we do not necessarily know that one is a subset of the other one. This is resolved by looking at common refinements:
Assume $P$ satisfies $L(f,P) > I^* - \varepsilon$ and $Q$ is an arbitrary partition. We need to show that then there is a refinement $Q^\prime$ of $Q$ such that $L(f,Q^\prime)\ge L(f,P)$ (and, consequently, $L(f,Q^\prime)>I^*-\varepsilon$).
For $Q^\prime$ one can choose the common refinement $R$: if $P=\{x_1,\ldots x_n \}$ and $Q=\{y_1,\ldots y_m \}$ then we just let $R = P\cup Q$. Since this is a refinement of both $P$ and $Q$ we have both $L(f,R)\ge L(f,P)$ as well as $L(f,R)\ge L(f,Q)$
Second edit: the original version was not correct:
For the other direction it suffices to show that if the function is integrable in the sense of the second definition then both $I_*$ and $I^*$ agree with the of the sums from the second definition. Since the reasoning is the same in both cases I'll just look at $I_*$.
So fix $\varepsilon >0$ and a given partition $P$ such that
$$|L - \sum_{i=1}^n f(\xi_i)\Delta x_i |< \varepsilon$$
if only the partition is fine enough.
Choose such a partition $P=\{x_0,\dots x_n\}$ and to $[x_{i-1},x_{i}]$ choose $\eta_i\in[x_{i-i},x_{i}]$ such that for
$m_i:=\inf \{ f(x):x\in [x_{i-1},x_i]\} $
we have $$0\le f(\eta_i)-m_i\le \frac{\varepsilon}{2n}$$
Then
\begin{eqnarray}
| L -\sum_{i=1}^n m_i \Delta x_i|
& = & |L- \sum_{i=1}^n f(\eta_i)\Delta x_i + \sum_{i=1}^n f(\eta_i)\Delta x_i
-\sum_{i=1}^n m_i\Delta x_i| \\
&\le & |L- \sum_{i=1}^n f(\eta_i)\Delta x_i| + \sum_{i=1}^n | f(\eta_i)
- m_i|\Delta x_i \\
& < & \frac{\varepsilon}{2} + \sum_{i=1}^n \frac{\varepsilon}{2n}=\varepsilon
\end{eqnarray}
If you 'see' that $0 <L -I_*< L -\sum_{i}m_i \Delta x_i$ then you are done here, otherwise it follows easily from the last estimate that the $\sum_i m_i \Delta x_i$ are, for any partition which is fine enough, $\varepsilon $ close to the fixed real number $L$, which of course implies that the $\sup$ over these sums exists and equals $L$ (here you need to use again the fact that you will approach the $\sup$, if it exists, if the width of the partitions goes to $0$).
The function $G$ takes the values $0$ and $1$ hence is bounded. Your argument is correct and proves that this function is not Riemann-integrable.
For $F$: fix a positive $\varepsilon$ and let $N$ be such that $1/N\lt\varepsilon$. Define the step function $f_1$ and $f_2$ by $f_1=0$ and $f_2=1/N$, except on $r_1,\dots,r_N$, where the value is $1$. Then $f_1\left(x\right)\leqslant F\left(x\right)\leqslant f_2\left(x\right)$ for any $x\in[0,1]$ and the Riemann integral of $f_2-f_1$ does not exceed $\varepsilon$. Remark that there is nothing specific about $1/n$; the function $x\mapsto\sum_{n=1}^{+\infty}\delta_n g_n\left(x\right)$ is Riemann-integrable on $[0,1]$ for any sequence $\left(\delta_n\right)_{n\geqslant 1}$ converging to $0$.
Best Answer
Since $x\in[0,1]$, $x^2 \le x \le 1$ so $$ m_{k} = \inf\left\{f(x)\ |\ x \in [x_{k-1}, x_{k}]\right\} = x_{k-1}^2 $$
Therefore, $$ L(f, P) = \sum_{k=1}^{n}m_k\, \Delta x_{k} = \sum_{k=1}^{n} x_{k-1}^2\, \Delta x_{k} $$
Consider the partition $P = \left\{0, \frac{1}{n}, \frac{2}{n}, \dots, 1\right\}$. Then \begin{align} L(f) &= \lim_{n\to \infty}\sum_{k=1}^{n} x_{k-1}^2\, \Delta x_{k} \\ &= \lim_{n\to \infty}\sum_{k=1}^{n} \left(\frac{k-1}{n}\right)^2\, \frac{1}{n} \\ &= \lim_{n\to \infty}\sum_{k=1}^{n} \frac{(k-1)^2}{n^3} \\ &= \lim_{n\to \infty}\frac{1}{n^3} \sum_{k=1}^{n} (k-1)^2 \\ &= \lim_{n\to\infty}\frac{1}{n^3}\sum_{k=1}^n\left(k^2-2k+1\right)\\ &= \lim_{n\to \infty}\frac{1}{n^3} \left[\left(\frac{n}{6} + \frac{n^2}{2} + \frac{n^3}{3}\right) - \left(n^2+n\right) + n\right] \\ &= \frac{1}{3} \\ \end{align}
Similarly, $U(f) = \frac{2}{3}$.
Since $U(f) \neq L(f)$, $f$ is not Riemann integrable on $[0, 1]$.
EDIT:
Definition: Let $P = \left\{x_0 = a, x_1, \dots, x_n = b\right\}$ be a partition of $[a, b]$ and let $\Delta x_i = x_i ā x_{iā1}$ for $1 \le i \le n$. Then the norm $||P||$ of $P$ is defined by $$ ||P|| = \text{max}\left\{\Delta x_i\ |\ 1 \le i \le n\right\} $$
Lemma: Let $f:[a, b] \to \mathbb{R}$ be a bounded function and $P$ be a partition of $[a, b]$, then $$ L(f) = \lim_{||P||\to 0} L(f, P) $$
Proof:
Given any $\epsilon > 0$, there exists a partition $Q = \{y_0 = a, y_1, y_2, \dots, y_n = b\} \subseteq [a, b]$ such that $$ L(f, Q) > L(f) - \frac{\epsilon}{2} $$
Let $M \in \mathbb{R}$ such that $|f(x)| < M$, $\eta = \text{min}\left\{\Delta y_i\ |\ 1 \le i \le n\right\}$ and $\delta < \text{min}\left\{\eta, \frac{\epsilon}{4(n-1)M}\right\}$.
Then for any partition $P = \{x_0 = a, x_1, x_2, \dots, x_m = b\} \subseteq [a, b]$ such that $||P|| < \delta$, \begin{align} \left|L(f, P \cup Q) - L(f, P)\right| &\le 2(n-1)M\delta \\ &< \frac{\epsilon}{2} \end{align}
Thus, \begin{align} L(f, P) &= \left[L(f, P) - L(f, P \cup Q)\right] + L(f, P \cup Q) \\ &> \left(-\frac{\epsilon}{2}\right) + L(f) - \frac{\epsilon}{2} \end{align} $\therefore ||P|| < \delta \,\Rightarrow\, L(f) - L(f, P) < \epsilon$