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.
Best Answer
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.