Proof using Fundamental Theorem of Calculus (Showing RHS = LHS)

$\newcommand{\d}[1]{\; \mathrm{d} #1}$ $\newcommand{\bb}[1]{\left( #1 \right)}$ $\newcommand{\sb}[1]{\left[ #1 \right]}$ We apply integration by parts and FTOC: \begin{align*} \int_0^x \bb{\int_0^u f(t) \d{t}} \d{u} &= \sb{u\int_0^u f(t) \d{t}}_{u=0}^{u=x} - \int_0^x u\underbrace{\bb{\int_0^u f(t) \d{t}}'}_{=f(u)\text{ by FTOC}}\d{u} \\ &= x\int_0^x f(t) \d{t} - \int_0^x uf(u)\d{u} \\ &= \int_0^x xf(u) \d{u} - \int_0^x uf(u)\d{u} \\ &= \int_0^x (x - u)f(u) \d{u} \end{align*}

Continuing from where you left off, integrate both sides wrt $u$ from $0$ to $x$ to get

$$\begin{equation}\begin{aligned} \int_{0}^{x}F'(u)du & = \int_{0}^{x}\left(uf(u) + \int_{0}^{u} f(t) dt\right) du \\ F(x) - F(0) & = \int_{0}^{x}uf(u)du + \int_{0}^{x}\left(\int_{0}^{u}\left(f(t) dt\right)\right) du \\ x\int_{0}^{x}f(t)dt - 0 & = \int_{0}^{x}uf(u)du + \int_{0}^{x}\left(\int_{0}^{u}\left(f(t) dt\right)\right) du \\ \int_{0}^{x}xf(u)du & = \int_{0}^{x}uf(u)du + \int_{0}^{x}\left(\int_{0}^{u}\left(f(t) dt\right)\right) du \\ \int_{0}^{x}\left(\int_{0}^{u}\left(f(t) dt\right)\right) du & = \int_{0}^{x}xf(u)du - \int_{0}^{x}uf(u)du \\ \int_0^x\left(\int_0^u f(t) dt\right) du & = \int_0^x f(u)(x-u) du \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Note that with going from the third to the fourth lines, since $t$ and $u$ are just dummy integration variables then, I used a substitution of $u = t$ in the LHS integral. This was so you can more easily see how that integral can be combined with the other integral in the second last line to then get the last line which is what you're trying to show.