= Solution
Retaining the nonlinear factor $f$ determines the order-$\epsilon$ forcing and the matching constants. Write $\delta=\sqrt\epsilon$. The inner leading equation and its <boundary condition> at $r=1$ give $f_0=C(1-r^{-1/2})$. Matching to the far-field value fixes $C=1$.
The tail $-r^{-1/2}$ becomes $-\delta x^{-1/2}$ under $x=\epsilon r$, so a regular expansion in integer powers of $\epsilon$ cannot suffice. This is a <matched expansion with square-root and logarithmic corrections>. Define
$$
E(x)=\int_x^\infty t^{-3/2}e^{-t}\,dt,
\qquad E'=-x^{-3/2}e^{-x},
\qquad E=2x^{-1/2}-2\sqrt\pi+2\sqrt x+O(x^{3/2}).
$$
The <exponential integral> here is normalized by the displayed definite integral. <Integration by parts> also gives $E(x)=2e^{-x}/\sqrt x-2\sqrt\pi\operatorname{erfc}\sqrt x$, using the <complementary error function>.
For the outer <asymptotic expansion>, put $F(x)=f(x/\epsilon)$ and
$$
F=1+\delta F_1+\delta^2F_2+\cdots,\qquad
L g=g''+\left(\frac{3}{2x}+1\right)g'.
$$
The transformed equation gives $LF_1=0$ and $LF_2=-F_1F_1'$, with both corrections vanishing at infinity. Matching the leading inner tail determines $F_1=-E/2$. Multiplying the equation for $F_2$ by its <integrating factor> yields
$$
\left(x^{3/2}e^xF_2'\right)'=\frac14 E(x).
$$
Fix the particular solution $G$ unambiguously by
$$
I(x)=\int_0^x E(s)\,ds
=\sqrt\pi\operatorname{erf}\sqrt x+xE(x),\qquad
G(x)=-\int_x^\infty t^{-3/2}e^{-t}I(t)\,dt.
$$
Then $(x^{3/2}e^xG')'=E$, $G(\infty)=0$, and $G$ has no $x^{-1/2}$ term at zero. The general outer correction is $F_2=\beta E+G/4$, where
$$
G(x)=4\log x+\mu+O(\sqrt x).
$$
The logarithm in this <asymptotic expansion> explains why the inner expansion needs an $\epsilon\log\epsilon$ term.
For the inner <asymptotic expansion>, all corrections larger than order $\epsilon$ solve the homogeneous leading equation and vanish at $r=1$. At order $\epsilon$, the forcing is
$$
f_2''+\frac{3}{2r}f_2'=-f_0f_0'
=-\frac12r^{-3/2}+\frac12r^{-2}.
$$
Integration gives the particular solution $1-\sqrt r+\log r$, which vanishes at $r=1$. Thus write
$$
f=(1+\delta A+\epsilon C\log\epsilon+\epsilon B)(1-r^{-1/2})
+\epsilon(1-\sqrt r+\log r)+o(\epsilon).
$$
Express this inner expansion in $x=\epsilon r$ and compare the terms that survive in the overlap $\epsilon\ll x\ll1$. The order-$\delta$ constant gives $A=\sqrt\pi$. At order $\epsilon$, the term $C\log\epsilon+\log r$ becomes $\log x+(C-1)\log\epsilon$, so $C=1$. The singular term $-\epsilon A/\sqrt x$ matches $2\epsilon\beta/\sqrt x$, giving $\beta=-\sqrt\pi/2$. Finally, the regular constants give $B+1=\pi+\mu/4$.
The requested expansions, including every correction through order $\epsilon$, are therefore
$$
\boxed{\begin{aligned}
f(r,\epsilon)={}&\left[1+\sqrt{\pi\epsilon}+\epsilon\log\epsilon
+\epsilon\left(\pi+\frac\mu4-1\right)\right](1-r^{-1/2})\\
&+\epsilon(1-\sqrt r+\log r)+o(\epsilon),\qquad r\ \text{fixed},\\
f(x/\epsilon,\epsilon)={}&1-\frac{\sqrt\epsilon}{2}E(x)
+\epsilon\left[-\frac{\sqrt\pi}{2}E(x)+\frac14G(x)\right]
+o(\epsilon),\qquad x>0\ \text{fixed}.
\end{aligned}}
$$
The inner formula satisfies the <boundary condition> at $r=1$ term by term; the outer formula satisfies the condition at infinity. Each formula has its own stated region of validity, and the <matched asymptotic expansion> relates them in their overlap.
The constant can also be made explicit. <Integration by parts> in the definition of $G$ gives $G=-EI-\int_x^\infty E(t)^2\,dt$, and $E(x)I(x)\to8$. Squaring the <complementary error function> expression for $E$ separates the integral into three elementary pieces:
$$
\int_x^\infty E^2
=4\int_x^\infty\frac{e^{-2t}}t\,dt
-8\sqrt\pi\int_x^\infty\frac{e^{-t}}{\sqrt t}\operatorname{erfc}\sqrt t\,dt
+4\pi\int_x^\infty\operatorname{erfc}^2\sqrt t\,dt.
$$
The first piece is $-4\log x-4\gamma-4\log2+o(1)$, where $\gamma$ is the <Euler--Mascheroni constant>. Substitution $u=\sqrt t$ makes the second integral tend to $\sqrt\pi/2$, because $(\operatorname{erfc}u)'=-2e^{-u^2}/\sqrt\pi$. <Integration by parts> in the third integral gives $1/2-1/\pi$. Consequently
$$
\boxed{\mu=4\gamma+4\log2+2\pi-4,\qquad
B=\frac{3\pi}{2}+\gamma+\log2-2.}
$$
Equivalently, defining $G$ by its convergent integral already determines the same matching constant without needing this closed form.
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