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
“Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural world.
The length of a writing pen is a scalar under rotations and Galilean transformations. This is a significant physical fact about our world.
But the fact that you can measure its length in various units is not significant, because units are inventions of humans, not of Nature. Physicists never say that the length of a writing pen “transforms” because you can choose to measure it in different length units. Different units such as inches and centimeters for a particular physical quantity like length do not have any physical significance at all.
Going back to your original question, the difference between a scalar and a vector under rotations should now be obvious: a scalar is a single number that stays the same under a rotation, while a vector is a directed quantity that requires three numbers to describe it, and under rotations these numbers transform into linear combinations of each other, as specified by the relevant rotation matrix.
Under any other transformation group, the distinction between scalars and vectors is similar.