I think it is helpful to think about these concepts on an arbitrary manifold. In that setting, there is no such thing as a position vector, because the manifold may not be a vector space; positions are specified instead by local coordinates.
The only reason we can identify positions with vectors in Euclidean space and move them around is that (1) Euclidean space is a vector space, and (2) the space carries an affine connection, which supplies a notion of parallel transport. You can't even "move around" tangent vectors on an arbitrary manifold unless you have a connection (though every Riemannian manifold comes with a natural one, the Levi-Civita connection); and in addition, as Ted says in his answer, you can't imagine those tangent vector glued "wherever you like" unless the tangent bundle is trivializable. The tangent bundle on a vector space is always trivializable.
A very useful example to keep in mind is polar coordinates in the Euclidean plane. Here a position is specified by $(r,\theta)$ or, especially in physics and engineering, a "position vector" $r\mathbf{\hat{r}}$. The former is the manifold point of view; the latter exploits the fact that we are laying curvilinear coordinates on what, at bottom, is a vector space. But you quickly learn this thing called a position vector is not really a position vector, because the vector $\mathbf{\hat{r}}$ changes from point to point. It doesn't live in the underlying space; rather, it lives in the tangent bundle, and the tangent spaces change from point to point. We are only able to think of it as a bona fide position vector because we imagine uprooting it from its tangent space and gluing it instead to the tangent space at the origin. But, to repeat, in a general manifold, the underlying space isn't necessarily a vector space and doesn't necessarily have a zero vector.
In general, to specify a position we just gives its local coordinates, and that is that: we don't need a vector to specify a position. You might wonder, then, where velocity vectors come from, because we're so used to getting velocity vectors by differentiating position vectors. But velocity vectors in arbitrary manifolds don't come from differentiating position vectors. They come from partial differentiation in directions specified by the local coordinates. More precisely, given local coordinates $(x_1, x_2)$, we represent a velocity vector as the operator $a\frac{\partial}{\partial x_1}+b\frac{\partial}{\partial x_2}$. (This is part of the reason, I think, many people struggle with the abstract definition of a tangent vector. For years they have been allowed to think of a velocity vector as the derivative of a position vector.)
You can see how the two points of view diverge when you ask a physicist or engineer to tell undergraduates how to represent velocity or acceleration vectors in planar polar coordinates. Most likely, they will draw a nice picture showing why the (time) derivative (along some curve) of $\mathbf{\hat{r}}$ is $\dot{\theta}\mathbf{\hat{\theta}}$. But the picture only works if you can parallel transport the tangent basis vectors $\mathbf{\hat{r}}$ and $\mathbf{\hat{\theta}}$ at a later point back along the curve to the tangent space at the initial point -- and it is exactly this seemingly innocent feature of the picture that requires an affine connection in general. What the calculation is really doing is exploiting the standard Euclidean connection to specify the covariant derivative / Christoffel symbols of planar polar coordinates.
The thing is, you might want to get some topology in the picture. In fact, if you do not, you can choose any bijection between the sphere and a $\mathbb R$-vector space, and you end up with a structure of vector space on your "sphere" (by transporting the structure). My point is, there exist such $\varphi$, but what you really want is not for $ \varphi_O$ to be only bijective : if your space already has a shape, you want it to be a homeomorphism.
And there is no homeomorphism between the sphere and a $\mathbb R$-vector space (for example because a vector space is contractible - you can shrink it continuously into a point - whereas the sphere is not ; you can look that up in any basic course of algebraic topology)
Best Answer
The inequality $a^Tx=a^T(\theta x_1+(1-\theta)x_2)\leq \theta b+(1-\theta)b=b$ holds if and only if when $0\leq \theta \leq 1$. If suppose, $\theta <0 $ the value of $(1-\theta)$ becomes negative. We cannot really draw any conclusion about the inequality between L.H.S and R.H.S, i.e., $a^T(\theta x_1+(1-\theta)x_2)$ and $\theta b+(1-\theta)b$.
Hope this helps.