Your intuition is correct. When we denote a tensor by its elements, say $g_{\mu\nu}$, we really mean $g_{\mu\nu}(x^\sigma)$, where $x^\sigma$ denotes the functional dependence on the coordinates.
So yes, when we contract a tensor, say :
$$ J_{\mu}=J^\nu g_{\mu\nu} $$
We really mean : $$ J_{\mu}(x^\sigma)= J^\nu(x^\sigma)g_{\mu\nu}(x^\sigma) $$ But we almost never write it, because it is redundant. However, when we do a change of coordinates, that's when it becomes actually important to at least have in mind what's happening. Let's denote
$$J^{\mu'}_{\nu}=\frac{\partial x'{}^{\mu}}{\partial x^\nu}$$
As well as :
$$J^{\mu}_{\nu'}=\frac{\partial x^{\mu}}{\partial x'{}^\nu}$$
If you haven't used the "primes on the indices" notation don't worry, it's just a neat trick to make less mistakes and remember formulas easier.
Then, we usually use the transformation rule :
$$V'^{\mu'}=V^{\nu}J^{\mu'}_{\nu}$$
Which tells us how the components of the $V^\mu$ transform into $V'{}^\mu$ in the new coordinates. But here, we have to be careful when we restore the x dependence. To be correct, the equation reads :
$$V'^{\mu'}(x'^{\sigma})=V^{\nu}(x^{\sigma})J^{\mu'}_{\nu}(x^{\sigma})$$
Notice how transforming the element $V^{\nu}(x^{\sigma})$, gives us $V'^{\mu'}(x'^{\sigma})$ which is evaluated at the transformed coordinates location. Of course, both $x'^{\sigma}$ and $x^{\sigma}$ denote the same physical point on the manifold, expressed in different coordinates.
To answer the last question about the <PERSON>, yes, the contraction is understood to be performed at each point in spacetime. To re-iterate, in full detail you have :
$$R(x^\sigma)=R_{\mu\nu}(x^\sigma) g^{\mu\nu}(x^\sigma)$$
Then one might ask oneself : why then do we say that the <PERSON> scalar is invariant under change of coordinates ? It can't be, since it may depend on the coordinates. 818
And this is somewhat true. If we apply the fully detailed change of coordinates formula, for the <PERSON> scalar (or any other scalar) we get :
$$ R'(x'^{\sigma})=R(x^{\sigma}) $$
and this is what we mean really when we say that a scalar tensor is "constant" under reparametrisation.
Let us take an easy example. Consider spherical coordinates $(r,\theta,\phi)$, and assume that the metric is such that the <PERSON> scalar is :
$$R = r \cos(\theta)$$
Now say we change to cartesian coordinates. If we apply the formula for scalars, we have that
$$R'(x,y,z)=R(r,\theta,\phi)=r\cos(\theta)$$
Now, on the LHS we want something that depends on $(x,y,z)$. That's when the formula for the change of coordinates kick in. We need to express $r=r(x,y,z)$, and $\theta=\theta(x,y,z)$. Luckily I conveniently chose the formula, and we can immediately see that $r\cos(\theta)=z$. Therefore :
$$R'(x,y,z)=z$$
Thus, the <PERSON> scalar expressed in both coordinates is $R(r,\theta,\phi)=r\cos(\theta)$ and $R'(x,y,z)=z$. Now we can see a bit better why we say they are "constant". 418