As mentioned by Choi, the inequality is true when $A$ and $B$ commute (since $A$ and $B$ can be simultaneously diagonalized).

Using Rayleigh quotient we can see that $\left|\sqrt{\lambda_{1}\left(A+B\right)}-\sqrt{\lambda_{1}\left(B\right)}\right|\leq\sqrt{\lambda_{1}\left(A\right)}$ holds. But unfortunately the counterpart is not true for $\lambda_{2}$.

Could you explain how you got the inequality? Hope we will get some clue from Seva's work.

EDITED ANSWER

Salle's answer is very instructive for me.
I would like to sketch here an elementary proof of the inequality $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ he gave above.

Noticing that $Tr(\sqrt{A})=\sqrt{Tr(A)+2\sqrt{det(A)}}$ and $det(A+B)\leq det(A)+det(B)$, for any positive definite $A$ and $B$ in dimension $2$, one applies $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$ twice and then get the inequality.

For your edited question, again I've only checked the case where $A$ and $B$ commute, and the answer is yes. But I failed to decipher the subtlety arises in the general case. Hope we will see a conclusive answer soon.

I guess you are considering B as a fixed vector and A a perturbation, which makes the inequality interesting.