Let
$$
x_n=\sum_{k=0}^nf\left(\frac{k}{n}\right)-n\int_0^1f(x)dx
$$
We will use the following result:
Lemma If $g:[0,1]\to\mathbb{R}$ is a continuously differentiable function. Then
$$
\frac{g(0)+g(1)}{2}-\int_0^1g(x)dx= \int_{0}^1\left(x-\frac{1}{2}\right)g'(x)dx.
$$
Indeed, this is just integration by parts:
$$\eqalign{
\int_{0}^1\left(x-\frac{1}{2}\right)g'(x)dx
&=\left.\left(x-\frac{1}{2}\right)g(x)\right]_{x=0}^{x=1}
-\int_0^1g(x)dx\cr
&=\frac{g(1)+g(0)}{2}-\int_0^1g(x)dx
}$$
Now applying this to the functions
$x\mapsto f\left(\frac{k+x}{n}\right)$ for $k=0,1,\ldots,n-1$ and adding the resulting inequalities we obtain
$$
x_n-\frac{f(0) +f(1)}{2} = \int_0^1\left(x-\frac{1}{2}\right)H_n(x)dx\tag{1}
$$
where,
$$
H_n(x)=\frac{1}{n}\sum_{k=0}^{n-1}f'\left(\frac{k+x}{n}\right)
$$
Clearly for every $x$, $H_n(x)$ is a Riemann sum of the function continuous $f'$, hence
$$
\forall\,x\in[0,1],\quad\lim_{n\to\infty}H_n(x)=\int_0^1f'(t)dt
$$
Moreover, $| H_n(x)|\leq\sup_{[0,1]}|f'|$. So,
taking the limit in $(1)$ and
applying the Dominated Convergence Theorem, we obtain
$$
\lim_{n\to\infty}\left(x_n-\frac{f(0) +f(1)}{2}\right)=
\left(\int_0^1f'(t)dt\right)\int_0^1\left(x-\frac{1}{2}\right)dx=0.
$$
This proves that
$$
\lim_{n\to\infty}x_n=\frac{f(0) +f(1)}{2}
$$
And consequently
$$
\lim_{n\to\infty}\left(\sum_{k=\color{red}{1}}^nf\left(\frac{k}{n}\right)-n\int_0^1f(x)dx\right)=\frac{f(1)-f(0)}{2}
$$
This is your problem:
But that says this function is not riemann integrable.
You correctly found that the upper and lower sums for this one partition are not equal (assuming your arithmetic is right - I didn't check). But when you look at other partitions you will find different bounds. When you consider all the partitions you'll find that the smallest number larger than all the lower sums is the same as the largest number smaller than all the upper sums. That's the definition of Riemann integrability.
Best Answer
The statement is false.
Take $f(x) = x$ then we have $$\int_0^1 f(u) du = \frac{1}{2}$$ and \begin{align*}\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}+\frac{1}{2n}\right) &= \frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{k}{n}+\frac{1}{2n}\right) \\\ &= \frac{1}{n}\left(\sum_{k=0}^{n-1}\frac{k}{n}+\sum_{k=0}^{n-1}\frac{1}{2n}\right) \\\\ &= \frac{1}{n}\left(\frac{n-1}{2} + \frac{1}{2}\right) \\\\ &= \frac{1}{n}\cdot\frac{n}{2} = \frac{1}{2}\end{align*}
hence:
$$\left|\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}+\frac{1}{2n}\right)-\int_{0}^{1}f\left(u\right)du\right| = \left|\frac{1}{2} - \frac{1}{2}\right| = 0\leq\frac{\varepsilon}{n}.\tag{1}$$ for all $\varepsilon > 0, n\in\mathbb{N}$
Although $f(x) = x$ is not positive on $[0,1]$ consider that a shift upwards doesn't change anything (it's added as a constant to the LHS and to the integral as well so it will cancel out).
So each function $f(x) = x + c$ with $c>0$ contradicts your claim.