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 $0$).
Under definition (1) or (2) we can show that a function $f$ cannot be both unbounded and Riemann integrable.
This can be shown by producing an $\epsilon > 0$ such that for any real number $A$, no matter how fine the partition, there is a Riemann sum with
$$|S(f,P) - A| > \epsilon$$
Given any partition $P$, since $f$ is unbounded, it must be unbounded on at least one subinterval $[x_{j-1},x_j]$ of P. Using the reverse triangle inequality we have
$$|S(f,P) - A| = \left|f(t_j)(x_j - x_{j-1}) + \sum_{k \neq j}f(t_k)(x_k - x_{k-1}) - A \right| \\ \geqslant |f(t_j)|(x_j - x_{j-1}) - \left|\sum_{k \neq j}f(t_k)(x_k - x_{k-1} - A \right|$$
Since $f$ is unbounded on $[x_{j-1},x_j]$, choose a partition tag $t_j$ such that
$$|f(t_j)| > \frac{\epsilon + \left|\sum_{k \neq j}f(t_k)(x_k - x_{k-1}) - A \right|}{x_j - x_{j-1}},$$
and it follows that no matter how fine the partition $P$ we have
$$|S(f,P) - A| > \epsilon.$$
Thus, when $f$ is unbounded, it is impossible to find $A$ such that for every $\epsilon > 0$ and sufficiently fine partitions, the condition $|S(f,P) - A| < \epsilon$ holds. We can always select the tags so that the inequality is violated.
Best Answer
Your definition is
where $\sum(f,P)$ means the Riemann sum of $f$ with respect to $P$. The alternative definition is
We would like to show one implies the other. First
$(1) \implies (2)$ Suppose $(1)$ holds. Since refinements can only decrease the mesh, and $\delta$ depends on $\epsilon$, the claim follows: for each $\epsilon>0$ take a partition $P_\epsilon$ with mesh $\delta'<\delta$ given by the above. Then any refiniment will have mesh at most $\delta'$ which will be less than $\delta$, and hence $(2)$ will hold.
$(2)\implies (1)$ This one is the tricky one. Here you can find a proof
An easy characterization of Riemann integrability is the following, which doesn't requiere that we know what the value of integral is:
Moreover, this is equivalent to $$\overline{\int_a^b} f=\underline{\int_a^b}f$$
so if you're able to prove the above and evaluate any upper or lower integral, you're done.
In fact, this applies to the Riemann Stieljes integral whenever the integrator $\alpha$ is montone.