According to my book:
The Taylor expansion for a function of two variables $f(x,y)$ around a stationary point $(a,b)$ has the appearance $$f(a+h,a+k) = > f(a,b)-\frac{1}{2}(Ah^2+2Bhk+Ck^2)+(h^2+k^2)^{3/2}B(h,k), \quad (1)$$
where $B(h,k)$ is a bounded function and $A=f''_{xx}(a,b), \ > B=f''_{xy}(a,b), \ C=f''_{yy}(a,b)$.
Using this with $(a,b)=(1,0)$ I get
$$f(1,0) = \sqrt{1+1+0}=\sqrt{2}\\ f'_x=f'_y=-\frac{1}{2\sqrt{1+x+y}} \\ f''_{x}=f''_{xy}=f''_{yy}=-\frac{1}{4(1+x+y)^{3/2}}$$
$$\implies A=B=C = -\frac{1}{4(1+1+0)^{3/2}}=-\frac{1}{8\sqrt{2}}.$$
Plugging it into (1) gives
\begin{array}{lcl} f(1+h,0+k) & = & \sqrt{2}-\frac{1}{16\sqrt{2}}(h^2+2hk+k^2) + (h^2+k^2)^{3/2}B(h,k) \\ & = & \sqrt{2}-\frac{1}{16\sqrt{2}}(h+k)^2 + (h^2+k^2)^{3/2}B(h,k) \\ \end{array}
Correct answer: $$\sqrt{2}+\color{green}{\frac{1}{2\sqrt{2}}(h+k)}-\frac{1}{16\sqrt{2}}(h+k)^2 + (h^2+k^2)^{3/2}B(h,k)$$
Where did the green term come from?
The Taylor expansion around a stationary point $(a,b)$ looks as given in text underlined in skin tone. Such a point is characterized by $f_x(a,b)=f_y(a,b)=0$; hence in this case there are no linear terms in the Taylor expansion. But the point $(1,0)$ is not a stationary point of your $f$ since $$f_x(x,y)=f_y(x,y)={1\over 2\sqrt{1+x+y}}\ ,$$ which is $={1\over2\sqrt{2}}\ne0$ at $(1,0)$ (you have a wrong sign here). This value produces the linear term $${1\over2\sqrt{2}}(h+k)$$ in the Taylor expansion of $f$ at $(1,0)$.