It says $f(x)$ is well-approximated by $L(x)$, at least near $x=0$. More precisely, near $0$, $f(x)=L(x)+ E(x)$, where $E(x)$ is an error term which is much smaller than $|x-x_0|$ when $x$ is near $x_0$.
How do we say the error term is much smaller than $|x-x_0$? By saying that $\dfrac{E(x)}{x-x_0}$ is near $0$ when $x$ is close to $x_0$. Another way of saying this is to say that $\frac{E(x)}{x-x_0}=\delta(x),$ where $\delta(x)\to 0$ as $x\to 0$. That can be rewritten as $E(x)=(x-x_0)\delta(x),$ and since $f(x)=L(x)+E(x)$, we get exactly the expression in the definition.
Remark: Much more informally, it turns out that $L(x)$ has the desired property precisely if $y=L(x)=a(x-x_0)$ is the tangent line to $y=f(x)$ at $x=x_0$. So $a=f'(x_0)$. The tangent line to $y=f(x)$ "kisses" $y=f(x)$ at $x=x_0$. If you look at $y=f(x)$ under very high magnification, near $x=x_0$ the curve will look like a straight line. Which straight line? The tangent line, the line $y=L(x)$.