With this definition, the separatrices are exactly the differentiable curves $g:I\to\mathbb R^2$ such that $g(I)\subseteq\mathbb R\times\{0\}$. Thus, $g(I)$ can be any interval of the $x$-axis. There are more restrictive definitions of this notion, though.
First let me address what your various textbooks say.
Given a manifold $M$, to say that a vector field $f$ is a mapping from $M$ to $TM$ is incomplete. Let me use $T_p M$ to denote the fiber of $TM$ over the point $p \in M$, in other words $T_p M$ is the tangent space of $M$ at $p$. A vector field on a manifold $M$ is not just any old mapping from $M$ to $TM$. Instead, a vector field on $M$ is a mapping $f : M \to TM$ such that for each $p \in M$ we have $f(p) \in T_p M$, in other words $f(p)$ is a vector in the tangent space at $p$.
In the special case where $M = \mathbb{R}^n$, if you keep all of this notation in mind, then your two textbooks are saying essentially the same thing. The tangent bundle in this case is a product $T \mathbb{R}^n = \mathbb{R}^n \times \mathbb{R}^n$, and the tangent space at each point $p\in \mathbb{R}^n$ has the form
$$T_p \mathbb{R}^n = \{(p,v) \,|\, v \in \mathbb{R}^n\}
$$
where the vector operations on the vector space $T_p \mathbb{R}^n$ are defined by simply ignoring the $p$ coordinate, i.e. $(p,v) + (p,w) = (p,v+w)$ and similarly for scalar multiplication. Because of this, there is a canonical isomorphism between vector fields expressed as functions
$$f : \mathbb{R}^n \to \mathbb{R}^n
$$
and vector fields expressed as functions
$$g : \mathbb{R}^n \to T \mathbb{R}^n
$$
This canonical isomorphism is given by the formula $g(p)=(p,f(p))$.
Regarding your last sentence, perhaps there may be elementary expositions of dynamical systems that restrict attention to $M=\mathbb{R}^n$, but the full theory of dynamical systems considers manifolds in all their full and general glory, and in this theory it is not sufficient to consider $\mathbb{R}^n$. Dynamical systems on spheres, on toruses, and on all kinds of manifolds are important.
I would also point out that it is misleading to say that a dynamical system on a manifold $M$ is a vector field. What is important in dynamical systems is not the vector field in particular, but its integral curves and their behavior over long time spans.
Best Answer
No. Any zero of $f(y)$ is an equilibrium, so the hypothesis that the only equilibria are $0$ and $1$ forces $f(y) < 0$ in $(0, 1)$. This in turn implies that for any $0 < \epsilon < 1/2$ we have $f(y) < 0$ on the closed interval
$I_\epsilon = [\epsilon, 1 - \epsilon], \tag 1$
and since $I_\epsilon$ is compact, $f(y)$ attains a global maximum $m < 0$ on $I_\epsilon$; then for
$y(t_0) = y_0 \in I_\epsilon, \tag 2$
$y(t)$ obeys
$\dot y = f(y) \le m < 0 \tag 3$
on $I_\epsilon$; therefore, $y(t)$ satisfying (3) will reach the value $\epsilon$ within time
$\Delta t = \displaystyle \int_{y_0}^\epsilon \dfrac{dy}{f(y)} = -\int_\epsilon^{y_0} \dfrac{dy}{f(y)} \le -\int_\epsilon^{y_0} \dfrac{dy}{m} = -\dfrac{y_0 - \epsilon}{m} = \dfrac{\epsilon - y_0}{m}; \tag 4$
since this holds for every $0 < \epsilon < 1/2$, we see that for every $y_0 \in (0, 1)$, $y(t)$ becomes arbitrarily small for large enough $t$; but this implies
$\displaystyle \lim_{t \to \infty} y(t) = 0. \tag 5$
The result is false if we remove the condition that $f(y)$ have only two equilibria in $[0, 1]$; consider
$f(y) = y \left (y - 1 \right ) \left ( y - \dfrac{1}{2} \right )^2, \tag 6$
and set
$y(0) = \dfrac{3}{4}; \tag 7$
$f(y)$ satisfies the requisite criteria, but
$\displaystyle \lim_{t \to \infty} y(t) = \dfrac{1}{2}, \tag 8$
which may be proved in a manner similar to the above.