Prove that $f(x) leq K cdotexp(Lcdot int_a^x g(t)dt)$

This is a particular case of Gronwall's inequality:

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Let $alpha$ and $betageq0$ be differentiable and continuous functions on $I:=[a,infty)$ respectively. If $x$ is a function on $I$ such that $$begin{align} x(t)leq alpha(t) + int^t_abeta(s) x(s),dstag{1}label{gr-cond} end{align}$$ then $$ x(t)leq alpha(t) + int^t_a alpha(s)beta(s)expBig(int^t_sbeta(r),drBig),ds $$ If in addition $alpha$ is nondecreasing then, $$ x(t)leqalpha(t)expBig(int^t_abeta(s),dsBig) $$

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Set $h(t)$ to be the right hand side of~eqref{gr-cond}. By the fundamental theorem of Calculus $$ dot{h}(t)=dot{alpha}(t) + beta(t)x(t)leqdot{alpha}(t)+beta(t)h(t). $$ That is, $$ begin{align} dot{h}(t)-beta(t)h(t) leq dot{alpha}(t)tag{2}label{two} end{align} $$ As in solving linear differential equations of first order, we may multiply both sides of $eqref{two}$ By the integrating factor $$expBig(-int^t_abeta(r),drBig)$$ to obtain $$ left(expBig(-int^t_abeta(r),drBig)h(t)right)' leq dot{alpha}(t)expBig(-int^t_abeta(r),drBig) $$ Integrating over $[a,t]$ gives $$ begin{align} expBig(-int^t_a beta(r),ddrBig),h(t)&leq alpha(a)+ int^t_adot{alpha}(s)expBig(-int^s_abeta(r),drBig),ds end{align} $$ Solving for $h$ gives $$ begin{align} h(t)&leq alpha(a)expBig(int^t_abeta(r),drBig)+ int^t_adot{alpha}(s)expBig(int^t_sbeta(r),drBig),dstag{4}label{gr-pre-by-parts} end{align} $$ Applying integration by parts to the second integral on the right-hand side leads to $$ x(t)leq h(t)leq alpha(t) + int^t_aalpha(s)expBig(int^t_sbeta(r),drBig),ds. $$ If $alpha$ is non--decreasing, then $dot{alpha}geq0$ and, since $betageq0$,~eqref{gr-pre-by-parts} reduces to $$ begin{align} x(t)leq h(t)&leq alpha(a)expBig(int^t_abeta(r),drBig)+ int^t_adot{alpha}(s)expBig(int^t_abeta(r),drBig),ds\ &= alpha(a)expBig(int^t_abeta(r),drBig)+ expBig(int^t_abeta(r),drBig)Big(alpha(t)-alpha(a)Big)\ &leq alpha(t) expBig(int^t_abeta(r),drBig) end{align} $$

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In your case, $alpha(t)equiv K$ and $beta(t)=L g(t)$

It is Gronwall's inequality, but I will give you a proof for your version:

Define $$h(x)=int_{a}^{x}f(t)g(t)dt$$, then take derivative of it w.r.t. $x$ and by the assumption given: $$h'(x)=f(x)g(x)-f(a)g(a)leq(K+Lh(x))g(x)-f(a)g(a)$$ $$frac{h'(x)+f(a)g(a)}{K+Lh(x)}leq g(x)$$ Then integrate w.r.t. $t$ from $a$ to $x$ and note that $h(a)=0$, then $$RHS=int_{a}^{x}g(t)dt$$ and $$LHS=int_{h(a)}^{h(x)}frac{h'(t)}{K+Lh(t)}dh=frac{1}{L}big(ln|K+Lh(x)|-ln|K+Lh(a)|big)=frac{1}{L}lnbig|frac{K+Lh(x)}{K}big|$$ Take exponents on both sides, then multiply $K$ on both sides, and by assumption, $$f(x)leq K+Lh(x)leq Kexpbig(Lint_{a}^{x}g(t)dtbig)$$ Done.