I assume that you made a typo and the choice for the input should be
$$
v = \ddot{y}_d - k_1\,e - k_2\,\dot{e},
$$
so using $\ddot{y}_d$ instead of $\ddot{y}$, which acts as a feedforward term. Such feedforward term makes sure that when the error is zero it will remain zero for any $y_d$ which is at least twice differentiable.
The choice for the input-output linearization gives $\ddot{y} = v$. So if we now look at the error dynamics one gets
\begin{align}
\ddot{e} &= \ddot{y} - \ddot{y}_d \\
&= v - \ddot{y}_d \\
&= \ddot{y}_d - k_1\,e - k_2\,\dot{e} - \ddot{y}_d \\
&= - k_1\,e - k_2\,\dot{e}
\end{align}
Coming up with this $v$ is common choice in linear control. Namely as stated before $\ddot{y}_d$ acts as feedforward. The remaining two terms act as state feedback, for example after applying the feedforward, so define $v = \ddot{y}_d + w$, the error dynamics can also be written as
$$
\dot{z} =
\underbrace{
\begin{bmatrix}
0 & 1 \\ 0 & 0
\end{bmatrix}
}_A z +
\underbrace{
\begin{bmatrix}
0 \\ 1
\end{bmatrix}
}_B w,
$$
with $z = \begin{bmatrix}e & \dot{e}\end{bmatrix}^\top$. Now by using state feedback $w = -K\,z$ we get the closed loop dynamics $\dot{z} = (A - B\,K)\,z$, which decays exponentially to zero if $A - B\,K$ is Hurwitz. Choosing a $K$ can be done with things like pole placement or LQR, but it can be shown that $A - B\,K$ is always Hurwitz when $k_1,k_2>0$, with $K = \begin{bmatrix}k_1 & k_2\end{bmatrix}$.
Lets say one would choose $v = \ddot{y} - k_1\,e - k_2\,\dot{e}$, plugging this into $\ddot{y}$ would give $k_1\,e + k_2\,\dot{e} = 0$. However, $v$ is chosen to be equal to $\ddot{y}$ so solving for $v$ would imply solving $v = v - k_1\,e - k_2\,\dot{e}$. This equation is only true when $k_1\,e + k_2\,\dot{e} = 0$ but you are not able to choose what $e$ and $\dot{e}$ are at any given moment. And if that equation would be true, then all values for $v$ would satisfy it. So either way it does not lead to a very sensible result.
Best Answer
The proposed control law makes the resulting dynamics linear. Namely, when substituting the control law in the dynamics one can factor out $a$ which gives
$$ \dot{a} = -\frac{k_p}{K_2}(a^*-a) - \frac{k_i}{K_2} \psi. $$
By using that $\dot{\psi}=a^*-a$ and that $a^*$ is constant, it follows that the second derivative of $\psi$ with respect to time would be $\ddot{\psi}=-\dot{a}$. Substituting the expression for $\dot{a}$ in yields
$$ \ddot{\psi} = \frac{k_p}{K_2} \dot{\psi} + \frac{k_i}{K_2} \psi, $$
which is linear and can be shown to be stable if $\frac{k_p}{K_2},\frac{k_i}{K_2}<0$.