According to the text:
f is a continuously differentiable function such that f(a,b)=0 for some point (a,b). $f'(a,b)$ is denoted by $A$.
F is defined as $$F(x,y)=(f(x,y),y)$$
To obtain the F'(a,b), after some calculation, we have $$ F(a+h,b+k) - F(a,b) = (A(h,k),k)+(r(h,k),0) $$ , where $r$ is the remainder that occurs in the definition of $f'(a,b)$
then the author had the result that
$F'(a,b)$ maps $(h,k)$ to $(A(h,k),k)$.
According to the definition of derivative, my calculation show that
$$ F'(a,b)(h,k) = F(a+h,b+k)-F(a,b)-R(h,k)= (A(h,k),k)+(r(h,k),0)-R(h,k) $$ ,where $R$ is the remainder that occurs in the definition of $F'(a,b)$
I don't figure out how the last two terms were canceled out. In other word, how do we know the following equality is true?
$$ (r(h,k),0)=R(h,k) $$
The point is that the error term $R(h,k)$ caused by linear approximation $$F(a+h,b+k)= F(a,b)+T(h,k)+R(h,k)$$ should satisfy $$ \lim_{(h,k)\to (0,0)}\frac{|R(h,k)|}{\sqrt{|h|^2+|k|^2}}=0. $$ If there exist a linear transformation $T$ such that $R(h,k)$ satisfies the above conditions, then such $T$ is unique and we say that $F$ is differentiable at $(a,b)$ and write $F'(a,b)=T$.
So, to prove the assertion that $F'(a,b)$ sends $(h,k)$ to $(A(h,k),k)$, i.e. $F'(a,b)(h,k)=(A(h,k),k)$, what we need to show is that $$ \lim_{(h,k)\to (0,0)}\frac{|(r(h,k),0)|}{\sqrt{|h|^2+|k|^2}}=\lim_{(h,k)\to (0,0)}\frac{|r(h,k)|}{\sqrt{|h|^2+|k|^2}}=0. $$ But this is obvious from the fact that $r(h,k)$ is an error term from another linear approximation $$f(a+h,b+k)=f(a,b)+A(h,k)+r(h,k)$$ of a differentiable map $f$ at $(a,b)$. Thus we conclude that $$F'(a,b)(h,k)=(A(h,k),k).$$