Let $V_1,\dots,V_n$ be be i.i.d. random variables from $V\sim U(0,a), \; a \gt0$ and $$f_V(x)=f(x) \cdot a \cdot I_{[0,1]}(x)$$
Based on my notes:
$$\mathbb{E_V}(f_V(V))=\int_{0}^{a}f_V(x)\cdot \frac{1}{a}dx=\int_{0}^{1}f(x)\cdot a\cdot \frac{1}{a}dx=\int_{0}^{1}f(x)dx=\mathbb{E_U}(f(U))$$
I can't understand the contribution of $\frac{1}{a}$ on first integral.
The density of $V$ is $g_V(v) = \frac{1}{a} \mathbb{1}_{(0,a)}(v)$, so the expected value of a function of $V$ by definition is
\begin{align*} E_V(f_V(V)) = \int_{\mathbb{R}}f_V(v) g_V(v) dv = \int_{\mathbb{R}}f_V(v) \frac{1}{a}\mathbb{1}_{(0,a)}(v) dv = \int_0^af_V(v) \frac{1}{a} dv \end{align*}
The rest of what you've written seems correct, assuming $a > 1$, or else the integral you have on the right should be taken over the region $(0,a)$.