I can easily construct an example of smooth tensor field over a manifold whose "rank" changes depending on the point. My idea relies upon the following elementary proposition.
I stress that the notion of "rank" used here is that introduced within the original question and not the standard one.
Proposition. Consider a $n$-dimensional real vector space $V$ and let $e_1,e_2 \in V$ be a pair of linearly independent vectors. The "entangled" tensor $$e_1\otimes e_1 + k e_2 \otimes e_2$$ is of "rank" $2$ necessarily if $k\neq 0$, otherwise it is of "rank" 1.
PROOF. Since $e_1$ and $e_2$ are linearly independent there is a basis $\{e_i\}_{i=1,\ldots,n}$ of $V$ containing both them.
Assume $k\neq 0$. If the tensor were fo rank 1, i.e.,
$$e_1\otimes e_1 + k e_2 \otimes e_2= u\otimes v\quad (1)$$
as we have $u= \sum_i u^i e_i$ and $v= \sum_j v^je_j$, (1) would entail:
$$e_1\otimes e_1 + k e_2 \otimes e_2= \sum_{i,j=1}^nu^iv^j e_i\otimes e_j\quad $$ and so:
$$0= (u^1v^1-1) e_1\otimes e_1 + (u^2v^2-k) e_2\otimes e_2 + u^1v^2 e_1\otimes e_2 + u^2v^1 e_2\otimes e_1 + \sum_{i,j >2} u^iv^j e_i\otimes e_j\qquad (2)$$
Since, in turn, $\{e_i\otimes e_j\}_{i,j=1,\ldots,n}$ is a basis of the space of tensors $V \otimes V$, (2) implies, in particular, that:
$$u^1v^1= 1\:, \:\:u^2v^2= k\:,\:\: u^1v^2=u^2v^1=0$$
multiplying together the first two conditions we have:
$$u^1v^1u^2v^2 = k$$
whereas the remaining ones entail:
$$u^1v^2u^2v^1 = 0$$
which are in contradiction unless $k=0$. QED
So consider a smooth (Hausdorff) manifold of dimension $n$ and a smooth function $\chi: M \to \mathbb R$ which constantly attains the value $1$ in an open set $U\subset M$ and smoothly vanishes before reaching the boundary of another open set $U' \supset U$. Taking $U$ and $U'$ small enough we can always assume that $U'$ is equipped by local coordinates $x^1,\ldots, x^n$. Under these hypotheses define the smooth tensor field $\Xi$ on $M$, vanishing outside $U'$:
$$\Xi(p) := \chi(p)\left(\frac{\partial}{\partial x^1}|_p \otimes \frac{\partial}{\partial x^1}|_p + k(p) \frac{\partial}{\partial x^2}|_p \otimes \frac{\partial}{\partial x^2}|_p\right) $$
where $k: M \to \mathbb R$ is a smooth function that vanishes somewhere in $U$ but not everywhere.
The tensor field $\Xi$ is smooth, well defined over the whole $M$, but it changes its "rank" three times: $0$ outside $U'$, $1$ and $2$ inside $U$, depending on the choice of $k$.
Though this example is completely mathematical, I think that with some further elaboration, some physical meaning could be given to my example, at least when the coordinate system is defined in the whole manifold of dimension $4$ and Lorentzian. The tensor is symmetric and we can assume that $x^1$ is a temporal Minkowskian coordinate so that $\partial_{x^1}$ could define the rest frame of some continuous body and $\Xi$ its stress-energy tensor. Assuming that this system interacts with some external system even the conservation low $\nabla_a\Xi^{ab}= J_{ext}^b$ can be imposed in order to have a non constant function $k$.
