$\newcommand{\de}{\delta}$You have the bounds 
\begin{equation}
	P(|X_i-m_i|\ge \de_i m_i]\le2e^{-\de_i^2 m_i/3}
\end{equation}
for $i=1,2$ and $\de_i>0$, where 
\begin{equation}
	m_i:=EX_i. 
\end{equation}
Clearly, these bounds are useful only if $m_i>0$ for $i=1,2$, which will be henceforth assumed. Assume also that $0<\de_i<1$ for $i=1,2$. 

Then on the event 
\begin{equation}
\begin{aligned}
	A&:=\{|X_1-m_1|<\de_1 m_1,|X_1-m_1|<\de_2 m_2\} \\ 
	&=\{m_1(1-\de_1)<X_1<m_1(1+\de_1),m_2(1-\de_2)<X_2<m_2(1+\de_2)\}
\end{aligned}
\end{equation}
we have $X_1>0$ and $X_2>0$, which implies that 
\begin{equation}
	Y = \frac{X_1-X_2}{X_1+X_2}
\end{equation}
is increasing in $X_1$ and decreasing in $X_2$, so that the event  
\begin{equation}
B:=\Big\{\frac{m_1(1-\de_1)-m_2(1+\de_2)}{m_1(1+\de_1)+m_2(1-\de_2)}	<Y<\frac{m_1(1+\de_1)-m_2(1-\de_2)}{m_1(1-\de_1)+m_2(1+\de_2)}\Big\}
\end{equation}
occurs. That is, $A\subseteq B$ and hence $1-P(B)\le1-P(A)$. 

So, by the union bound, we get the concentration result: 
\begin{equation}
	1-P(B)\le 2e^{-\de_1^2 m_1/3}+2e^{-\de_2^2 m_2/3}. 
\end{equation}
One can now (quasi-)optimize the choices of $\de_1$ and $\de_2$, depending on one's objectives.