A proof without measure theory: Note that if $f'\in R[a,b],$ then $\int_a^b f' = f(b)-f(a).$ Proof: Briefly, the MVT shows
$$f(b)-f(a) = \sum_{k=1}^{n} (f(x_k) - f(x_{k-1})) = \sum_{k=1}^{n} f'(c(k,n))\Delta x_k \to \int_a^b f'.$$
Now for $f\in R[0,1],$ define
$$S_n(f) = \sum_{k=1}^{n}f(k/n) - n\int_0^1f.$$
Lemma: If $f'\in R[0,1],$ then
$$\mid S_n(f) \mid \le \int_0^1|f'|.$$
Proof: We can write
$$S_n(f) = n\sum_{k=1}^{n}\int_{(k-1)/n}^{k/n}(f(k/n) - f(t))\,dt
=n\sum_{k=1}^{n}\int_{(k-1)/n}^{k/n}\int_{t}^{k/n}f'(s)\,ds\,dt.$$
Now take absolute values, replacing $f'(s)$ by $|f'(s)|$ and $t$ by $(k-1)/n.$ We get
$$|S_n(f)| \le \sum_{k=1}^{n}\int_{(k-1)/n}^{k/n}|f'(s)|\,ds= \int_0^1|f'|.$$
For the main result, assume $f'\in R[0,1].$ We use two facts. First, as you said, the result holds if $f'$ is continuous. Second, any Riemann integrable function can be approximated in the "Riemann norm" by continuous functions. So let $\epsilon>0.$ Then there exists $g$ continuous on $[0,1]$ such that $\int_0^1|f'-g| < \epsilon.$ Define $G(x) = \int_0^x g.$ We know the result holds for $G.$ So
$$S_n(f) - (f(1)-f(0))/2 = S_n(f-G) + [S_n(G)- (G(1)-G(0))/2] + [(G(1)-G(0))-(f(1)-f(0)]/2.$$
The last summand above equals $(1/2)\int_0^1(g-f').$ Taking absolute values and using the above results, we get
$$\limsup_{n\to \infty} |S_n(f) - (f(1)-f(0))/2| \le \epsilon + 0 + \epsilon/2.$$
Since $\epsilon$ is arbitrary, we have the desired limit.