(1) It's because it's the form of your equation. Note Culomb's law for point charges
$$
U(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{r} \tag{1}
$$
is also only valid for a point source located at the origin. The coordinate free version is
$$
U(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \frac{q}{|\mathbf{r - r'}|}
$$
where $\mathbf{r}$ is the field point and $\mathbf{r'}$ is the vector from the origin to the charge $q$. Indeed, this is the most general form of the potential of a point charge that reduces to (1) in the special case that $\mathbf{r'}=0$. That is,
$$
U(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \frac{q}{|\mathbf{r}|} = \frac{1}{4\pi\epsilon_0} \frac{q}{r}
$$.
Why does a point charge have to be located at the origin?
It doesn't. Note that moving the point charge off the origin amounts to making the variable substitution $\mathbf{r} \to \mathbf{r-r'}$ in the differential equation. This doesn't change the nature of the solution, since $\mathbf{r'}$ is constant, and so we may write with impunity that
$$
\varphi(\mathbf{r})=a\,\frac{e^{\mu_{\gamma}|\mathbf{r-r'}|}}{|\mathbf{r-r'}|}+b\frac{e^{-\mu_{\gamma}|\mathbf{r-r'}|}}{|\mathbf{r-r'}|}
$$
is the correct "coordinate free" version. Note that that your boundary condition now changes to $\lim_{|\mathbf{r-r'}|\to \infty} \varphi < \infty$ since the whole idea is that we are getting arbitrarily far away from the point charge.
(2) If the field generated by a point source didn't go to zero at infinity, it will either (a) go to a constant or (b) diverge.
(a) If it goes to a constant, then we can just take that to be our "zero point" for potential, since only changes in potential matter.
$\bullet$ That is, truthfully, the most general form is
\begin{equation}
\Delta\varphi(r)=a\,\frac{e^{\mu_{\gamma}r}}{r}+b\frac{e^{-\mu_{\gamma}r}}{r}
\end{equation}
where $\Delta\varphi = \varphi(r) - \varphi(r_0)$ where $r_0$ is some reference point. We just usually take $r_0$ to be at $\infty$. Note that if we impose the condition then that $\varphi(\infty) = 0$ then we have that $\varphi(r_0) = 0$ if we take $r_0$ to be at $\infty$, and we can therefore just write $\varphi(r)$ instead of $\Delta\varphi$, since with this choice $\Delta\varphi = \varphi(r)$.
(b) If the actual physical field diverges at infinity, then that just simply would not describe a point particle. So we impose this condition on physical grounds.
this condition leads to the conclusion $a=0$ (Why?).
The equation
\begin{equation}
\varphi(r)=a\,\frac{e^{\mu_{\gamma}r}}{r}+b\frac{e^{-\mu_{\gamma}r}}{r}
\end{equation}
is the general solution to a second order differential equation, and so $a,b$ are arbitrary constants that must be determined from the boundary conditions. One of your conditions is
$$ \lim_{r\to\infty}\varphi (r) \to \rm{finite} $$
So applying this yields
\begin{align*}
\lim_{r\to\infty}\varphi(r)&=a\lim_{r\to\infty}\frac{e^{\mu_{\gamma}r}}{r}+b\underbrace{\lim_{r\to\infty}\frac{e^{-\mu_{\gamma}r}}{r}}_{\searrow 0}\\
&=a\lim_{r\to\infty}\frac{e^{\mu_{\gamma}r}}{r} \\
& \to \infty
\end{align*}
The term in the last line diverges, so the only chance we have to satisfy the boundary condition is to eliminate the term all together by taking $a=0$.
