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.
Your small angle approximation is wrong and your Jacobian linearization is incomplete.
If $x_1 \approx 0$ then $\cos(x_1) \approx 1$ and so the system
$$
\begin{align}
\dot{x_1} &= x_2 \\
\dot{x_2} &= \cos(x_1) \\
\end{align}
$$
behaves close to $(0,0)$ approximately like
$$
\begin{align}
\dot{x_1} &= x_2 \\
\dot{x_2} &= 1 \\
\end{align}
$$
If you put this in matrix form:
$$
\begin{pmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} \begin{pmatrix}
x_1 \\
x_2
\end{pmatrix} + \begin{pmatrix}
0 \\
1
\end{pmatrix}
$$
This is the "small angle approximation" and as you can see this is essentially the same as the Jacobian linearization. Remember that the Jacobian linearization of a nonlinear system $\dot{x} = f(x)$ at $x_0$ is
$$
\dot{x} = f(x_0) + A(x - x_0)
$$
where $A$ is the Jacobian matrix of $f$ at $x_0$. If you insert your system and your point you get:
$$
\begin{pmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
-\sin(0) & 0
\end{pmatrix} \begin{pmatrix}
x_1 - 0 \\
x_2 - 0
\end{pmatrix} + \begin{pmatrix}
0 \\
\cos(0)
\end{pmatrix}
$$
which is just:
$$
\begin{pmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} \begin{pmatrix}
x_1 \\
x_2
\end{pmatrix} + \begin{pmatrix}
0 \\
1
\end{pmatrix}
$$
so the same as the small angle approximation.
Your mistake was that you neglected the fact that you linearized at a non-equilibrium point. Usually we linearize at an equilibrium and then $f(x_0) = 0$, so the constant term can be neglected.
But you linearize at $(0, 0)$ which is not an equilibrium point of your system because $\dot{x_2} \neq 0$ at $x_1 = 0$, so the system is not at rest at the origin.
Best Answer
By selecting $x_1=x$ and $x_2=\dot{x}$ you can write
$$\begin{align*} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= k_1 x_1 (1 - |x_1|) + k_2 u \end{align*}$$
Now the fixed points are $(0, 0), (1, 0), (-1, 0)$. The Jacobian around $(1, 0)$ (as an example) is
$$ \begin{bmatrix} 0 & 1 \\ k_1(1 - 2 x_1) & 0 \end{bmatrix} |_{x_1=1} = \begin{bmatrix} 0 & 1 \\ -k_1 & 0 \end{bmatrix} $$