The command
Grad[V[r],{r,th},"Polar"]
will give you an answer in polar coordinates, not in cartesian coordinates. That means that the results
{V'[r],0}
represents the gradient in polar coordinates
\begin{equation}
{\rm grad} (V(r)) = V'(r) e_r + 0 e_\theta
\end{equation}
If you want, you can then transform this back to cartesian coordinates, but that's usually not the point in using an alternative coordinate system, you want to describe it in the new system.
EDIT: may be this needs some background clarification. From vector/tensor analysis, depending on the transformation rule of the coordinates you have to consider a lot. In Mathematica, if you use a general function depending on all variables, you will get the general coordinates of the gradient in the respective system. Example: for polar coordinates
Grad[f[r, th], {r, th}, "Polar"]
you will get the general form for polar coordinates
{D[f[r,th],r],D[f[r,th],th]/r}
As commented in the question by Lukas, for a function NOT depending on $\theta$, the second component vanishes. This has nothing to do with the $\theta$ position of the point considered in the gradient. The gradient is a new vector, which can be represented in respect to any chosen basis. If you want to derive the gradient of a general SCALAR function $f$
\begin{equation}
{\rm grad}f
= \frac{\partial f}{\partial r} e_r + \frac{\partial f}{\partial \theta} \frac{1}{r} e_\theta
\end{equation}
consider this:
Transformation of coordinates, with Cartesian coordinates $x^i$ in respect to a position INdependent orthonormal basis $\{e_i\}, e_i\cdot e_j = \delta_{ij}$ ($\delta_{ij}$ is the Kronecker symbol) and polar coordinates $\{r,\theta\} = \{u^1,u^2\}$
\begin{equation}
\begin{bmatrix}
x^1 \\ x^2
\end{bmatrix}
=
u^1
\begin{bmatrix}
\cos(u^2) \\ \sin(u^2)
\end{bmatrix} \ , \qquad
\begin{bmatrix}
u^1 \\ u^2
\end{bmatrix}
=
\begin{bmatrix}
\sqrt{(x^1)^2+(x^2)^2} \\ \arctan(x^2/x^1)
\end{bmatrix}
\end{equation}
The gradient of a general function $f$ is originally defined in respect to Cartesian coordinates as
\begin{equation}
{\rm grad}f
= \sum_{i=1}^2\frac{\partial f}{\partial x^i}e_i
\end{equation}
With the chain rule you can get this
\begin{equation}
{\rm grad}f
= \sum_{i=1}^2\frac{\partial f}{\partial x^i}e_i
= \sum_{i=1}^2\sum_{j=1}^2\frac{\partial f}{\partial u^j}\frac{\partial u^j}{\partial x^i}e_i
= \sum_{j=1}^2\frac{\partial f}{\partial u^j} \sum_{i=1}^2 \frac{\partial u^j}{\partial x^i}e_i
= \sum_{j=1}^2\frac{\partial f}{\partial u^j} g^j \ ,
\end{equation}
at what we defined the so called gradient basis vectos $\{g^j\}$
\begin{equation}
g^j = \sum_{i=1}^2 \frac{\partial u^j}{\partial x^i}e_i \ .
\end{equation}
The gradient basis vectors are position dependent. They are the dual vectors of the natural tangential basis vectors $g_j = \sum_{i=1}^2 \partial x^i/\partial u^j e_i$, i.e., $g_m \cdot g^n = \delta_{mn}$. For polar coordinates you get here
\begin{equation}
g^1
= \sum_{i=1}^2 \frac{\partial u^1}{\partial x^i}e_i
= \frac{x^1}{\sqrt{(x^1)^2+(x^2)^2}}e_1 + \frac{x^2}{\sqrt{(x^1)^2+(x^2)^2}}e_2
= \cos(u^2) e_1 + \sin(u^2) e_2 \ , \qquad
g^2
= -\frac{\sin(u^2)}{u^1}e_1 + \frac{\cos(u^2)}{u^1}e_2 \ .
\end{equation}
Sadly, these guys are, in general, nor orthogonal to each other, nor normalized. For the special case of polar coordinates, the are orthogonal, but not normalized. So if we want to describe our gradient, a vector, in respect to a orthonormal basis, we have to normalize at least the chosen basis vectors
\begin{equation}
{\rm grad}f
= \sum_{j=1}^2\frac{\partial f}{\partial u^j} g^j
= \sum_{j=1}^2\frac{\partial f}{\partial u^j} ||g^j|| \frac{g^j}{||g^j||}
= \sum_{j=1}^2\frac{\partial f}{\partial u^j} ||g^j|| g^{*j}
\end{equation}
with the normalized basis vectors
\begin{equation}
g^{*j} = \frac{g^j}{||g^j||} \ .
\end{equation}
At the end you get, that for a general function in polar coordinates, its gradient is equivalently described as
\begin{equation}
{\rm grad}f
= \sum_{i=1}^2\frac{\partial f}{\partial x^i}e_i
= \sum_{j=1}^2\frac{\partial f}{\partial u^j} ||g^j|| g^{*j}
= \frac{\partial f}{\partial u^1} g^{*1} + \frac{\partial f}{\partial u^2} \frac{1}{u^1} g^{*2}
= \frac{\partial f}{\partial r} e_r + \frac{\partial f}{\partial \theta} \frac{1}{r} e_\theta
\end{equation}
If you consider a function not depending on theta, $f(r,\theta) = V(r)$, then you get the result given above and what Mathematica also gives you. Mathematica does all this stuff internally, also for the gradients of higher-order tensors. You can use the gradient function of Mathematica for second-order tensors and you will get the answer using this and actually more stuff (Christoffel symbols of the second kind, ...).
