Let $f: [a,b] \rightarrow \mathbb R$ be a function defined on a closed interval $[a,b]$ of the real numbers, $\mathbb R$, and
$\mathcal P = \left \{[x_0,x_1],[x_1,x_2],\dots,[x_{n-1},x_n] \right \}$,
be partition of the closed interval $[a,b]$, where
$ a=x_0<x_1<x_2<\cdots<x_n=b $.
One, perhaps the, way of defining the Riemann upper sum is as follows.
$ U = \displaystyle{\sum_{i=1}^{n} f(x_i^*)\, \Delta x_i }$
where $ \Delta x_i=x_i-x_{i-1} $, $ x_i^*\in[x_{i-1},x_i]$, and $f(x_i^*) = \sup f([x_{i-1},x_i]) $ (that is, the supremum of $f$ over $[x_{i-1},x_i]$).
My question is how does this work for a function with a discontinuity at what would be it's maximum in a particular sub-interval $[x_{i-1},x_i]$; say the function $f$ has a `hole' at the point $c\in [x_{i-1},x_i]$?
For example, take $f(x) = -x^2 + 1$ on $[-1,1]$ with $(0,1)$ removed, and consider a partition with one of the sub-intervals $[-\epsilon, \epsilon]$, $\epsilon >0$. In this situation, what value do you pick for the $x^*$ value in the sub-interval? I mean, it's clear that you'd take a rectangle of height 1 for this bin, since that is the supremum of the set of values of $f$ on this interval, but what value do you take for $x^*$? Does it matter, since the supremum for the set doesn't depend on the chosen value for $x^*$?
I think my misunderstand is how to interpret choosing an $x^*_i \in [x_{i-1},x_i]$ when we are really using the supremum of the set of values of $f$ on $[x_{i-1},x_i]$. Does it matter which value we pick for $x_i^*$, since the supremum of the set is fixed?
It appears that you are having some confusion on Riemann sum, Upper Darboux sum, Lower Darboux sum.
Based on Apostol's Mathematical Analysis, I provide the following definitions which are pretty standard and followed in many other textbooks.
Let $[a, b] $ be a closed interval and let function $f:[a, b] \to\mathbb {R} $ be bounded on $[a, b] $. A partition $P$ of $[a, b] $ is a finite set of points from interval $[a, b] $ and necessarily includes the end points. Typically partition $P$ is written as $$P=\{x_0,x_1,x_2,\dots,x_n\} $$ where $$a=x_0<x_1<x_2<\dots<x_n=b$$ Let us define $$M_k=\sup\, \{f(x) \mid x\in[x_{k-1},x_k]\},k=1,2,\dots,n$$ and $$m_k=\inf\, \{f(x) \mid x\in[x_{k-1},x_k]\}, k=1,2,\dots,n$$ Since $f$ is bounded the numbers $M_k, m_k$ exist. Also since we are not given that $f$ is continuous, these values $M_k, m_k$ may or may not be attained by $f$.
The upper Darboux sum for $f$ over partition $P$ of $[a, b] $, denoted by $U(f, P) $, is defined as $$U(f, P) =\sum_{k=1}^{n}M_k(x_k-x_{k-1})$$ In a similar manner the lower Darboux sum $L(f, P) $ is defined as $$L(f, P) =\sum_{k=1}^{n}m_k(x_k-x_{k-1})$$ Riemann sums are a bit more complicated in the sense that they not only depend upon the partition, but also on a further chosen set of points called tags.
Let $t_1,t_2,\dots,t_n$ be points in $[a, b] $ such that $t_k\in[x_{k-1},x_k]$ for each $k$ and let $$T_P=\{t_1,t_2,\dots,t_n\}$$ The notation $T_P$ is used to emphasize that tag points are chosen based on a given partition.
A Riemann sum for $f$ over partition $P$ of $[a, b] $ with tag points in $T_P$, denoted by $S(f, P, T_P) $, is defined as $$S(f, P, T_P) =\sum_{k=1}^{n}f(t_k)(x_k-x_{k-1})$$
Notice that for any partition $P$ and any tag set $T_P$ we have $$L(f, P) \leq S(f, P, T_P) \leq U(f, P) $$ because $m_k\leq f(t_k) \leq M_k$. Also we can choose the tag points $t_k$ such that $f(t_k) $ is close to $M_k$ (or $m_k$) and hence any upper or lower Darboux can be well approximated by a suitable Riemann sum.
More formally (and try to prove it) if $\epsilon>0$ then we can choose tag sets $T_P, T'_P$ such that $$U(f,P) - \epsilon <S(f, P, T_P) \leq U(f, P), \\ L(f,P) \leq S(f, P, T'_P) <L(f, P)+\epsilon $$ The equality in $\leq $ of these relations given above may or may not occur. In the special case of continuous $f$ such an equality is possible for suitable choice of tags.
Thus you should not try to express an upper Darboux sum as a Riemann sum. They are different but related concepts and it may not be possible to express one as another.
Note: The term Riemann upper sum is not standard and most probably being used in place of the standard term upper Darboux sum.