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.,
\begin{equation*}
P_s(\infty-)=\infty,\quad Q_s(\infty-)=0,\quad Q_s>0\text{ on }(1,\infty). \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*}
Also, if $a\in(0,1/2)$, then the inequality in (3) implies $B>0$. So, $F(\infty-)\ne0$ -- unless $c_2=0$ and hence $F=0$.
Thus, there is no nonzero solution to your differential problem.