Fulton further specifies that the field $k$ is infinite. Fix a choice $(x_1,\ldots,x_{n+1})$ of homogeneous coordinates for $P$. Then $F_i(x_1,\ldots,x_{n+1})$ ($i\geqslant 0$) is a fixed element of $k$. Fulton asks that you consider $$G(\lambda)=F(\lambda x_1,\ldots,\lambda x_{n+1})=\sum_{i\geqslant 0}F_i(\lambda x_1,\ldots,\lambda x_{n+1})=\sum_{i\geqslant 0}\lambda^iF_i(x_1,\ldots,x_{n+1});$$ you already observed these equalities: they follow from the fact that $F_i$ is homogeneous of degree $i$. Now you may regard this as a polynomial in the one variable $\lambda$ whose coefficients are $F_0(x_1,\ldots,x_{n+1}),F_1(x_1,\ldots,x_{n+1}),\ldots$, so that $$G(T)=\sum_{i\geqslant 0}a_iT^i\in k[T],\;\;a_i=F_i(x_1,\ldots,x_{n+1})\in k.$$ By assumption, $F$ vanishes on all homogeneous coordinates of $P$: in terms of $G$, this means that $\lambda$ is a root of $G$ for all $\lambda\in k^*$. Then, it is a standard fact of algebra that a one-variable polynomial with coefficients in $k$ which vanishes at infinitely many elements of $k$ is the zero polynomial: it is the case here because $k$ is assumed to be infinite, so that $k^*=k\setminus\{0\}$ is also infinite. That $G$ is the zero polynomial means by definition that the coefficients of $G$, which are by definition the $a_i=F_i(x_1,\ldots,x_{n+1})$, are identically zero. Thus the $F_i$'s vanish at $(x_1,\ldots,x_{n+1})$ as required.
The trick is not to plug in some clever $\lambda$ to isolate the vanishing of the $F_i$'s. Indeed, this does not work in general. In particular, it follows from the proof that if $k$ is a finite field of order $d$ and if the degree of $F$ is larger than $d$, it may happen that $F$ vanish at all homogeneous coordinates of $P$ without this being the case for the individual $F_i$'s. Rather, you consider all the $\lambda$'s at once by putting them in a polynomial. This is quite standard to show that a finite family of elements of a field is identically zero.