I don't know what they meant by FLEA.
But $\Delta f$ means the change of the function in a short period of time. For example if the function is $f(x) = x^3 - 2x^2 + 3x + 1$ and we want to see the change if $f(x)$ from $x = 2$ and $x + h = 2.01$ That'd be $\Delta f(2.01) - f(2) = [(8.120601) - 2(4.0401) + 3(2.01) + 1] - [8 - 2*4 + 3*2 + 1] = .120601 - .0802 + .03 = .070401$ where $x = 2$ and $h = .01$.
Now the derivative of $f(x)$ is the formula for the rate of change of $f$ and the precise point $x$. So $f'(x) = 3x^2 - 4x + 3$ means at $x = 2$ the function is changing at a rate of $f'(2) = 3*4 - 4*2 + 3 =7$ units (in the "y-direction") for every unit (in the "x-direction").
Thus in $h$ units the function $f(x)$ should increase by $h*f'(x)$ units. So at $x = 2$ and $h = .01$ the function should increase by $0.01*7 = .07$. And indeed as we saw the function did increase by $0.070401$. That's close to $h*f'(x)$.
So ... now I think I do know what that mean be FLEA. $f'(x)$ is the rate of change at precisely $x$. $f(x)$ is constantly changing as is the rate of change. You can use the rate of change (the derivative) to approximate where the function will be at $x + h$ but as the function is changing there will be a slight error. This error is FLEA.
So $\Delta f = f(x + h) - f(x) = h*f'(x) + FLEA$. This is a fundamental formula for expressing changes in the value of $f$ in terms of the derivative/(intantaneous rat of change).
And in our example $\Delta f = f(2 + .01) - f(.01) = .070401 = (0.01)f'(2) + FLEA = 0.01*7 + FLEA = 0.07 + FLEA = 0.070401$. (FLEA = 0.000401).