I will abuse notation a little, but I hope you don't find it terrible. After all, I only abuse notation: not children!
Trick 1: Parametrize by Proper Length. We will pick for our affine parameter $\lambda=s$ the proper length. Then the stress energy tensor becomes
$$\tag{1}T^{\alpha\beta}(x)=m\int_{\gamma}u^{\alpha} u^{\beta}\frac{\delta^{(4)}\bigl(x,z(s)\bigr)}{\sqrt{|g|}}\,\mathrm{d}s$$
where $u^{\alpha}=\mathrm{d}x^{\alpha}/\mathrm{d}s$ and $g=\det{g_{\mu\nu}}$.
Trick 2: Covariant Derivative Trick. We can write
$$\nabla_{\mu}f^{\mu}=\frac{1}{\sqrt{|g|}}\partial_{\mu}(\sqrt{|g|}f^{\mu})$$
for arbitrary $f^{\mu}$.
Exercise: Using Poisson's notation (13.2), we have
$$\delta(x,x') = \frac{\delta^{(4)}(x-x')}{\sqrt{|g|}} = \frac{\delta^{(4)}(x-x')}{\sqrt{|g'|}}$$
and thus using our Covariant Derivative trick, find
$$\nabla_{\mu}\delta(x,x')=???$$
This will tell you that
$$\int u^{\alpha}u^{\beta}\nabla_{\alpha}\delta(x,x')\,\mathrm{d}s = \mbox{boundary terms}$$
and thus we can ignore it.
Remark 1. You are in error writing
\begin{alignedat}{1}\nabla_{\beta}T^{\alpha\beta} & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\delta_{4}\left(x,z\right)\right]d\lambda}=\\
& =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]\delta_{4}\left(x,z\right)d\lambda-m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\nabla_{\nu}\left[\delta_{4}\left(x,z\right)\right]d\lambda,}}
\end{alignedat}
This should have been a simple application of the product rule. That is, the minus sign should be a plus sign.
Remark 2. Why should we expect the right hand side of $\nabla_{\beta}T^{\alpha\beta}=0$? Well, because using Einstein's field equation it's $\nabla_{\beta}G^{\alpha\beta}$ and this is identically zero.
This is why we set
$$\tag{2}m\int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]\delta\bigl(x,z\bigr)\,\mathrm{d}\lambda=0.$$
...which is precisely the geodesic equation for a point particle as discussed in Poisson's article section 3.
Edit
We can rewrite (2) since ${g^{\alpha}}_{\beta}={\delta^{\alpha}}_{\beta}$ is the Kronecker delta. So
$$\tag{3}m\int_{\gamma}\nabla_{\nu}\left[\frac{\dot{z}^{\alpha}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\beta}}}\right]\delta\bigl(x,z\bigr)\,\mathrm{d}\lambda=0.$$
But if we pick the arclength as the parameter, this becomes simply
$$\tag{4}m\int_{\gamma}\nabla_{\nu}(u^{\alpha}u^{\nu})\delta\bigl(x,z\bigr)\,\mathrm{d}\lambda=0.$$
Great, but really is
$$\tag{5}\nabla_{\nu}(u^{\alpha}u^{\nu})=0?$$
Lets recall for a geodesic using arclength parametrization we have
$$u_{\mu}u^{\mu}=1\implies u_{\mu}\nabla_{\nu}u^{\mu}=0.$$
Thus (5), when contracted by a non-negative vector (say $u_{\alpha}$) becomes
\begin{alignedat}{1}u_{\alpha}\nabla_{\nu}(u^{\alpha}u^{\nu}) &= u_{\alpha}\underbrace{u^{\nu}\nabla_{\nu}u^{\alpha}}_{=0} + \underbrace{u_{\alpha}u^{\alpha}}_{=1}\nabla_{\nu}u^{\nu}\\
&=\nabla_{\nu}u^{\nu}\end{alignedat}
But this is a continuity-type equation (and if you use trick 2, it really resembles electromagnetism's continuity equation!).
Now we can go back, and by inspection we find
$$\nabla_{\nu}(u^{\alpha}u^{\nu})=\left(\begin{array}{c}\mbox{Geodesic}\\
\mbox{Equation}\end{array}\right)+\left(\begin{array}{c}\mbox{Continuity}\\
\mbox{Equation}\end{array}\right).$$
This is, of course, $\nabla_{\mu}T^{\mu\nu}$. Why should we expect it to be zero?
Well, if the Einstein field equations hold, then
$$G^{\mu\nu}-\kappa T^{\mu\nu}=0$$
and moreover
$$\nabla_{\mu}(G^{\mu\nu}-\kappa T^{\mu\nu})=0.$$
However, $\nabla_{\mu}G^{\mu\nu}=0$ identically thanks to geometry.
@Prahar is right, the variation of the Christoffel symbol is a tensor, even if the Christoffel itself is not. We have
$\delta \Gamma^\rho_{\mu\nu}=\frac{1}{2}\delta\bigg(g^{\rho\alpha}(2\partial_{(\mu}g_{\nu)\alpha}-\partial_\alpha g_{\mu\nu})\bigg)=\frac{1}{2}\delta g^{\rho\alpha}(2\partial_{(\mu}g_{\nu)\alpha}-\partial_\alpha g_{\mu\nu})+ \frac{1}{2}g^{\rho\alpha}(2\partial_{(\mu}\delta g_{\nu)\alpha}-\partial_\alpha \delta g_{\mu\nu})$
where $A_{(\mu\nu)}=\frac{1}{2}(A_{\mu\nu}+A_{\nu\mu})$. Using $\delta g^{\rho\alpha}=-g^{\rho\gamma}g^{\alpha\delta}\delta g_{\gamma\delta}$ we have:
$\delta \Gamma^\rho_{\mu\nu}=\frac{1}{2}g^{\rho\alpha}(2\partial_{(\mu}\delta g_{\nu)\alpha}-\partial_\alpha \delta g_{\mu\nu}-2\Gamma_{\mu\nu}^\beta\delta g_{\alpha\beta})$
The Christoffel then combines nicely with the standard derivative to give a covariant tensor (the other Christoffel symbols cancel each other)
$\delta \Gamma^\rho_{\mu\nu}=\frac{1}{2}g^{\rho\alpha}(2\nabla_{(\mu}\delta g_{\nu)\alpha}-\nabla_\alpha \delta g_{\mu\nu})$.
So to answer the original question, we finally have:
$\nabla_\mu V_\nu=\nabla_\mu \delta V_\nu-\frac{1}{2}g^{\rho\alpha}(2\nabla_{(\mu}\delta g_{\nu)\alpha}-\nabla_\alpha \delta g_{\mu\nu})A_\rho$
Remember that we did not assume anything on $V_\mu$. Depending on the problem, it is then possible to integrate by parts to isolate $\delta g_{\mu\nu}$ and obtain the energy momentum tensor.
Best Answer
Just be careful with what quantity depends on what argument, cf. above comment by user NowIGetToLearnWhatAHeadIs. Then it works like a charm:
$$\begin{align} \nabla^{(y)}_{\mu} T^{\mu\nu}(y) ~=~& \partial^{(y)}_{\mu} T^{\mu\nu}(y) ~+~\Gamma^{\mu}_{\mu\lambda}(y) T^{\lambda\nu}(y) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr ~=~& \frac{1}{\sqrt{-g(y)}}\partial^{(y)}_{\mu} \left(\sqrt{-g(y)}T^{\mu\nu}(y)\right) +\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr ~\stackrel{(2)}{=}~&\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau ~\dot{x}^{\nu}\dot{x}^{\mu}\partial^{(y)}_{\mu}\delta^4(y\!-\!x(\tau )) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr ~=~&-\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau ~\dot{x}^{\nu}\dot{x}^{\mu}\partial^{(x)}_{\mu}\delta^4(y\!-\!x(\tau )) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr ~=~&-\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau ~\dot{x}^{\nu} \frac{d}{d\tau}\delta^4(y\!-\!x(\tau )) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr \stackrel{\text{int. by parts}}{=}&~\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau ~\ddot{x}^{\nu} \delta^4(y\!-\!x(\tau)) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y)\cr &~-~\frac{m}{\sqrt{-g(y)}}\left[\dot{x}^{\nu}\delta^4(y\!-\!x(\tau))\right]_{\tau=\tau_i}^{\tau=\tau_f} \cr ~~~~~~\stackrel{(2)}{=}~&\frac{m}{\sqrt{-g(y)}} \int_{\tau_i}^{\tau_f} \!\mathrm{d}\tau\underbrace{\left\{\ddot{x}^{\nu}+ \Gamma^{\nu}_{\mu\lambda}(x(\tau))\dot{x}^{\mu}\dot{x}^{\lambda} \right\}}_{\text{geodesic eq.}}\delta^4(y\!-\!x(\tau ))\cr &~-~\frac{m}{\sqrt{-g(y)}}\left[\dot{x}^{\nu}\delta^4(y\!-\!x(\tau))\right]_{\tau=\tau_i}^{\tau=\tau_f}\cr \stackrel{\text{geodesic eq.}}{=}&~~-~\frac{m}{\sqrt{-g(y)}}\underbrace{\left[\dot{x}^{\nu}\delta^4(y\!-\!x(\tau))\right]_{\tau=\tau_i}^{\tau=\tau_f}}_{\text{source terms}}. \end{align}$$ The source terms naturally break the continuity equation (1) because they correspond to the creation & annihilation of energy-momentum of a particle. Away from creation & annihilation source terms, the continuity equation (1) should be satisfied, which then enforces the geodesic equation. $\Box$