First of all, your main equation contains $f(1)$ and therefore is not an ODE. Let us consider the ODE 
\begin{equation*}
	(r^2 - 2ar)f''(r) + 2(r-a) f'(r) - (4a + m(m+1))f(r) = p, \tag{1}
\end{equation*}
where $p$ is a real number; your equation corresponds to (1) with 
\begin{equation}
	p=-4af(1). \tag{2}
\end{equation}
The general real solution of (1) is given by 
\begin{equation*}
	f(r)=f_p(r):=
c_1 P_s\left(\frac{r}{a}-1\right)+c_2
   Q_s\left(\frac{r}{a}-1\right)-\frac{p}{4 a+m^2+m},
\end{equation*}
where $c_1$ and $c_2$ are arbitrary real constants; 
$P_s$ and $Q_s$ are, respectively, the Legendre functions of the first and second kinds whose values are real on $(1,\infty)$; and 
\begin{equation*}
	s:={\frac{1}{2} \left(\sqrt{4 m^2+4 m+16 a+1}-1\right)}>1. 
\end{equation*}
Obtaining now the root (say $p_*$) of the equation $p=-4af_p(1)$ (cf. (2)) for $p$ and substituting $p_*$ for $p$, we get the general solution $F$ of your main equation: 
\begin{equation*}
	F(r):=f_{p_*}(r)=
c_1 P_s\left(\frac{r}{a}-1\right)
+c_2 Q_s\left(\frac{r}{a}-1\right)+c_1 A+c_2 B,
\end{equation*}
where 
\begin{equation*}
A:=	\frac{4 a P_s\left(\frac{1}{a}-1\right)}{m (m+1)},\quad
B:=\frac{4 a Q_s\left(\frac{1}{a}-1\right)}{m(m+1)}.
\end{equation*}

According to [Sections 15.23 and 15.33 of Whittaker and Watson, 4th ed.][1],  
\begin{equation*}
P_s(\infty-)=\infty,\quad Q_s(\infty-)=0. \tag{3}	
\end{equation*}
So, the condition $F(\infty-)=0$ implies that $c_1=0$ and hence 
\begin{equation*}
	F(\infty-)=
c_2 B, 
\end{equation*}
which is in general nonzero. 

Thus, in general no solution to your differential problem exists. 


[1]: https://www.cambridge.org/core/books/course-of-modern-analysis/0702B9DE7A1F91437980EAE5F4F8B2A3