1.You're using a somewhat circuitous way to find bifurcation points. Most of the work is computing the equilibria of this differential equation,
$$
y = \frac{a \pm \sqrt{a^2- 4}}{2} ~~~~~ (*)
$$
Inspecting these forms, the parameter values $a = \pm 2$ correspond to the points at which equilibria are 'created' or 'destroyed', and so correspond to parameter values for which the qualitative aspects of the system change drastically. More specific to this one-dimensional problem, the phase line changes as $a$ crosses $\pm 2$ and remains the same elsewhere.
What you're doing, effectively, is solving for the parameter value(s) when these equilibria coincide; maybe in a more complicated system, it might be useful to pursue this method, but here you ought to be able to read off this behavior directly from $(*)$.
2.The phase line will depend on the value of $a$, and so a complete answer would include a phase line for each of the 5 cases: $a < -2, a = -2, -2 < a < 2, a = 2, a > 2$.
For each of $a < -2, a > 2$, there are two equilibria. I can tell you that in each case, one of them is a sink and other a source; you have to figure out which is which by writing
$$
\dot{y} = (y - \lambda_1) ( y - \lambda_2)
$$
where $\lambda_i$ are the equilibria given by $(*)$, and figuring out the sign of $\dot{y}$ for each of $y \in (-\infty, \lambda_1), (\lambda_1, \lambda_2), (\lambda_2, \infty)$.
For $-2< a < 2$ there are no equilibria, and the sign of $\dot{y}$ is positive for all $y$. What phase line corresponds to this situation? (Hint: it's pretty simple).
For $a = \pm 2$, there is a single equilibrium at $y = \pm 1$ that is neither a sink nor a source, because the sign of $\dot{y} = (y \mp 1)^2$ is always positive.
