Consider the affine plane curve defined by $F(X,Y)=XY$ and referred to simply as $F$. It is clear that $(0,0)$ is a double point of $F$ since if I decompose $F$ as a sum of its homogeneous parts, then I will simply get $F=F_2$ with $F_2=XY$, and thus by the definition of the multiplicity of a point I have that $(0,0)$ is a point of multiplicity two.
But let me replace $(0,0)$ with a point $(a,b)\ne (0,0)$. Then Fulton's book says that to determine the multiplicity of that point, I should consider $$F(X+a,Y+b)=(X+a)(Y+b)=ab+(aY+bX)+XY=F_0+F_1+F_2,$$ so again by definition $0$ is the multiplicity of $(a,b)$. But on the other hand $F_X(a,b)=2b$, which is never zero (I assume the field has characteristic zero) and so by a second definition this point is simple, that is, has multiplicity $1$. What do I misunderstand? Also, what are the tangent lines to $F$ at $(a,b)$ in this case? By definition they are the $L_i$ such that $F_m=\prod L_i^{r_i}$ with $m$ minimal, but in this case $m=0$, $F_0=ab$, and the equations of lines $a=0$, $b=0$, which cannot be because I asuumed $(a,b)\ne (0,0)$.
If you take a point $(a,b)\neq (0,0)$, you still need the point to be on the curve, that is you need $ab=0$ (in other words $a=0$ or $b=0$). If $ab\neq 0$, you are not on the curve (hence the multiplicity is zero as you said).
If $ab=0$, then $F(a+X,b+X)=0+(aY+bX)+XY$, so this is indeed a point of multiplicity 1 (unless $a=b=0$).
It is coherent with the derivative criterion. $F_X(a,b)=2b$ and $F_Y(a,b)=2a$. So either both of them is zero (in case $a=b=0$) and the multiplicity is $>1$. Or one of them is not zero and the multiplicity is 1. Note however that to apply the derivative criterion, you must take a point on the curve.
Finally the tangents are very easy to visualize. Note that the equation $XY=0$ defines an algebraic set which is the union of the two lines $X=0$ and $Y=0$, so if you are on a point $(a,0), a\neq 0$, you are on the $Y=0$ branch and the tangent at that point is again $Y=0$. You can also see this by computation : $F_1=aY+0X=aY$, so the tangent is given by the equation $aY=0$ which is equivalent to $Y=0$. (And similarly for the points of the form $(0,b)$).