The assumption tells you that $x_+\equiv R$ is a super-solution, and $x_-\equiv -R$ a sub-solution, of your problem. This means that $x_-''\ge f(t,x_-)$, $x_+''\le f(t,x_+)$, and $x_-\le0\le x_+$ at $t=0,1$.

It is a general fact that when a BVP for a second-order differential equation admits a sub- and a super-solution $x_\pm$, and if $x_-\le x_+$, then the problem admits a solution $x$ such that $x_-\le x\le x_+$.

Idea : Take $\lambda>0$ large enough that $x\mapsto g(t,x):=f(t,x)-\lambda x$ be non-increasing over $[-R,R]$, for every $t\in[0,1]$. Rewrite the problem as
$$x''-\lambda x=g(t,x).$$With $x(0)=x(1)=0$, this is equivalent to an integral equation
$$x(t)=-\int_0^1K_\lambda(t,s)g(s,x(s))ds=:(Tx)(t).$$
The kernel $K_\lambda$ is bounded and continuous. Let us define the convex subset $C$ of ${\mathcal C}(0,1)$ by the conditions
$$x(0)=x(1)=0,\qquad x_-\le x \le x_+.$$
Because of the monotonicity of $g$, and because $Tx_-\ge x_-$, $Tx_+\le x_+$, $T$ maps $C$ into itself. In addition, $T(C)$ is relatively compact (a consequence of Ascoli-Arzela). By Schauder fixed point theorem, it admits a fixed point $\bar x$. This $\bar x$ is a solution of your problem.