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 $0$).
Best Answer
For $a<b$, we define $\int_a^b f(x)\textrm{d}x$ as the supremum of the lower Darboux sums. On the other hand, we define $\int_b^a f(x)\textrm{d}x=-\int_a^b f(x)\textrm{d}x$ in order to make the identity
$$ \int_a^b f(x)\textrm{d}x+\int_b^c f(x)\textrm{d}x=\int_a^c f(x)\textrm{d}x $$ hold no matter the relation of $a,b$ and $c$.
Alternatively, when integrating from $a$ to $b,$ you observe a partition $a=x_0<x_1<...x_n<b$ and observe the Riemann sums $\sum_{j=1}^n f(\xi) (x_j-x_{j-1})$. The analogous construction when integrating from $b$ to $a$ would be to observe a partion $b=x_0>x_1>...>x_n=a$ and the corresponding Riemann sums $\sum_{j=1}^n f(\xi) (x_j-x_{j-1})$, and here, the $x_j-x_{j-1}$ is a negative increment. This is another way to intuit that this convention might be useful.