Well the question which you have mentioned is pretty easy. It is the converse which is bit hard to prove. Thus your accepted criteria of Riemann integrability is as follows:
(1) Let $f$ be bounded on $[a, b]$. Then $f$ is Riemann integrable on $[a, b]$ if for any $\epsilon > 0$ we can can find a partition $\mathcal{P}$ of $[a, b]$ such that $U(f, \mathcal{P}) - L(f, \mathcal{P}) < \epsilon$.
Thus for integrability check we need to find one partition $\mathcal{P}$ for every $\epsilon > 0$ which meets the above condition.
Now consider your question which asks to prove Riemann integrability of $f$ on $[a, b]$ provided that $f$ satisfies the following condition:
(2) For any $\epsilon > 0$ there is a number $\delta > 0$ such that for any partition $\mathcal{P}$ of $[a, b]$ with $||\mathcal{P}|| < \delta$, we have $U(f, \mathcal{P}) - L(f, \mathcal{P}) < \epsilon$.
Thus for any $\epsilon > 0$ we have a $\delta > 0$ which satisfies the condition written above. Let $n$ be a positive integer such that $0 < (b - a)/n < \delta$. Then we can see that the partition $\mathcal{P}$ defined by $$\mathcal{P} = \{x_{0}, x_{1}, x_{2}, \dots, x_{n}\},\,\, x_{i} = a + \frac{i(b - a)}{n}$$ has the norm $||\mathcal{P}|| = \dfrac{b - a}{n} < \delta$ and hence we have $$U(f, \mathcal{P}) - L(f, \mathcal{P}) < \epsilon$$ So for any $\epsilon > 0$ we have found a partition $\mathcal{P}$ which satisfies the accepted criteria $(1)$ for Riemann integrability. This proves that $f$ is Riemann integrable on $[a, b]$.
Note that the following converse is a bit difficult to prove.
(3) Let $f$ be bounded on $[a, b]$ and suppose that for every $\epsilon > 0$ there is a partition $\mathcal{P}$ of $[a, b]$ for which $$U(f, \mathcal{P}) - L(f, \mathcal{P}) < \epsilon$$ Then for every number $\epsilon > 0$ there is a number $\delta > 0$ such that for any partition $\mathcal{P}$ of $[a, b]$ with $||\mathcal{P}|| < \delta$ we have $$U(f, \mathcal{P}) - L(f, \mathcal{P}) < \epsilon$$
I did check the Pete Clark notes which you have linked in your post. The dicing lemma mentioned there is a slightly more general result than the result $(3)$ above and the proof is slightly more difficult so you may just use the simpler result I mentioned above and it is sufficient for our purpose here.
You can follow the proof in my blog post (taken from Tom Apostol's Mathematical Analysis). The basic idea is to choose a partition $\mathcal{P}_{\epsilon}$ such that $$U(f, \mathcal{P}_{\epsilon}) - L(f, \mathcal{P}_{\epsilon}) < \frac{\epsilon}{2}$$ and then let $N$ be number of points in partition $\mathcal{P}_{\epsilon}$. We can choose $\delta = \epsilon/(4KN)$ where $K = \sup\,\{|f(x)|\} + 1$. Now if $\mathcal{P}$ is any partition of $[a, b]$ with $||\mathcal{P}|| < \delta$ then we can split the expression $$\Delta(f, \mathcal{P}) = U(f, \mathcal{P}) - L(f, \mathcal{P}) = \sum_{i = 0}^{n}(M_{i} - m_{i})(x_{i} - x_{i - 1})$$ into two parts $S_{1}, S_{2}$ such that $S_{1}$ has the terms corresponding to the subintervals $[x_{i - 1}, x_{i}]$ which contain no points of the partition $\mathcal{P}_{\epsilon}$ and $S_{2}$ has terms corresponding to the remaining subintervals made by $\mathcal{P}$. Thus each subinterval in $S_1$ is contained in some subinterval created by $P_{\epsilon} $ and $S_2$ consists of remaining subintervals.
Clearly this will ensure that subintervals used in making $S_{1}$ are contained in subintervals made by partition $\mathcal{P}_{\epsilon}$. Hence we have $$S_{1} \leq U(f, \mathcal{P}_{\epsilon}) - L(f, \mathcal{P}_{\epsilon}) < \frac{\epsilon}{2}$$ For sum $S_{2}$ we can see that the subintervals used here do contain points of $\mathcal{P}_{\epsilon}$ and hence the number of such subintervals will not be greater than the number of points in $\mathcal{P}_{\epsilon}$. Also length of each subinterval is less than $\delta$. Thus we can see that $$S_{2} < 2K \cdot N \cdot \delta = \frac{\epsilon}{2}$$ Thus we have $$\Delta(f, \mathcal{P}) = U(f, \mathcal{P}) - L(f, \mathcal{P}) = \sum_{i = 0}^{n}(M_{i} - m_{i})(x_{i} - x_{i - 1}) = S_{1} + S_{2} < \epsilon$$
Update: Since the question has been updated I am adding some more details. We have to prove that
If $f$ is bounded on $[a, b]$ then $f$ is Riemann integrable on $[a, b]$ if and only if there for every $\epsilon > 0$ there is a number $\delta > 0$ such that if $\mathcal{P}$ is a partition of $[a, b]$ with $||\mathcal{P}|| < \delta$ then $$U(f, \mathcal{P}) - L(f, \mathcal{P}) < \epsilon$$
In view of statement $(1)$ in my answer this proposition is same as showing that the following two statements are equivalent:
Let $f$ be bounded on $[a, b]$
(A) For every $\epsilon > 0$ there is a partition $\mathcal{P}$ of $[a, b]$ such that $$U(f, \mathcal{P}) - L(f, \mathcal{P}) < \epsilon$$
(B) For every $\epsilon > 0$ there is a $\delta > 0$ such that for any partition $\mathcal{P}$ with $||\mathcal{P}|| < \delta$ we have $$U(f, \mathcal{P}) - L(f, \mathcal{P}) < \epsilon$$
We need to show that both $(A)$ and $(B)$ are equivalent i.e. $(A) \Leftrightarrow (B)$. Proof of $(B)\Rightarrow (A)$ is given after statement $(2)$. Proof of $(A) \Rightarrow (B)$ is given after statement $(3)$.
