$PT-, T-, P-$ transformations refer to subgroup of discrete transformations of the Lorentz group. They transform connected components of the Lorentz group between each other ($PT$ transformation transforms $L^{\uparrow}_{+}$ representation to $L^{\downarrow}_{+}$). In general, they can't be represented as the special case of rotation, which refer to subgroup of continuous transformations. You can't get some other connected component from orthochronous group by acting of any continuous transtormation's matrix on your representation. So, by nature, it can't be rotation.
It turns out that a vector field $\vec F(x, y, z)$ is best thought of as a local directional derivative which we'd write $\mathsf F= \vec F \cdot \nabla$, taking scalar fields $f(x,y,z)$ and producing new scalar fields $g(x,y,z)$ from them. When we start from this place of$$g(x,y,z) = \mathsf F[f](x,y,z) = F_x\frac{\partial f}{\partial x} + F_y\frac{\partial f}{\partial y} + F_z\frac{\partial f}{\partial z}$$ for scalar fields $F_{x,y,z}(x, y, z)$, and we define that $\hat g(x, y, z) = g(-x, -y, -z)$ and $\hat f(x, y, z) = f(-x, -y, -z)$, as the parity transform asks us to perform, we will find that the chain rule tells us that $\partial_x \hat f = -\partial_x f,$ and we get one big minus sign on the right hand side. We can correct for this if we define that $\hat F_{\alpha}(x, y, z) = -F_\alpha(-x, -y, -z).$ So there is both a minus sign "inside" and "outside" the parentheses; the inner one comes from the fact that $\vec F$ needs to be using parity-transformed inputs and the outer one comes from the fact that this reflection actually changed the direction of $\vec F$ itself.
In spherical coordinates (I will take $0\le \theta < 2\pi$ as azimuthal while $0\le \varphi\le\pi$ comes down from the North Pole) both of these ideas get a little complicated; we instead have $$g(r,\theta,\varphi) = \mathsf F[f](r, \theta, \varphi) = F_r \frac{\partial f}{\partial r} + F_\theta \frac1{r\sin\varphi} \frac{\partial f}{\partial \theta} + F_\varphi\frac1{r}\frac{\partial f}{\partial\varphi}.$$Our parity transform needs to map latitudes to opposite latitudes $\varphi\mapsto\pi-\varphi$ and needs to also rotate a point 180 degrees about the globe, $\theta\mapsto\theta + \pi.$ Under this mapping actually $\sin\varphi\mapsto\sin\varphi,$ so there's a sort of "double negative" in this $\sin$ term and we therefore find that: $$\begin{array}{ll}
F_r(r,\theta,\varphi) &\mapsto~ {+F_r(r, \theta + \pi, \pi - \varphi)},\\
F_\theta(r,\theta,\varphi) &\mapsto~ {+F_\theta(r, \theta + \pi, \pi -
\varphi)},\\
F_\varphi(r,\theta,\varphi) &\mapsto~ {-F_\varphi(r, \theta + \pi, \pi -
\varphi)}.\end{array}$$Another way to interpret this is to just think about the unit vectors at a given point in space; remember that when we're using curvilinear coordinates these unit vectors vary over space itself, so unlike $\hat x$, the unit vector in the $x$-direction which is the same at all points in a Cartesian space, the unit vector $\hat r$ changes to point away from the origin at all points in space. In fact one can work out that these unit vectors are:$$\begin{array}{ll}
\hat r &=~ {+\hat z}~\cos\varphi + \hat x~ \sin\varphi \cos\theta + \hat y\sin\varphi\sin\theta\\
\hat \theta &=~ {-\hat x}~\sin\theta + \hat y~ \cos\theta\\
\hat \varphi &=~ {-\hat z}~\sin\varphi + \hat x~ \cos\varphi \cos\theta + \hat y~\cos\varphi\sin\theta
\end{array}$$
When we reflect space our starting $\hat r$ gets mapped to the reflected $\hat r$; they both are unit vectors pointing away from the origin. Similarly our starting $\hat \theta$ gets mapped to the reflected $\hat\theta$; they both are unit vectors pointing "West". But when we reflect $\hat \varphi$, which points "South" everywhere on the sphere, we get that the reflected vector points "North", which is opposite of its new local-unit-vector. So that component, and only that component, must pick up the minus sign.
Best Answer
The parity transformation needs to have a determinant of $-1$. In all dimensions, it is safe to define it as $(x_1, \dots, x_d) \mapsto (-x_1, x_2, \dots, x_d)$ where only one co-ordinate is flipped. Let's call this the good parity transformation because it generalizes well to any $d$.
If $d$ is odd, $(x_1, \dots, x_d) \mapsto (-x_1, \dots, -x_d)$ is also valid and this seems to be what you learned first. I will call this the bad parity transformation because in even $d$ it would just be a rigid rotation with determinant $1$.
The point is that the good and bad parity transformations in odd dimension differ by a rigid rotation. I.e. something with determinant $1$. This also explains why there was no loss of generality in reflecting $x_1$ instead of $x_2$ for example in the good one. So you can quotient $O(d)$ by any of these and get the same subgroup $SO(d)$ as a result.