I don't follow the Taylor expansion of the ODE $C_1\dot v(t)+ C_2v(t)+v^2(t)-F_m(t)$
Let $$f(F_m,v,\dot{v})=C_1\dot v+ C_2v+v^2-F_m=0. \tag{1}$$ Define an equilibrium point $v(t)=v_0$, let $\dot v=0$ and solve for $F_m=F_0$ $$F_0=C_2v_0+v_0^2$$ Thus, the equilibrium point where $f(F_0,v_0,0)=0$ is $(F_0,v_0)=(C_2v_0+v_0^2,v_0)$.
Taylor expansion of $(1)$ about the point $(F_0,v_0)$ yields \begin{align} f(F_m,v,\dot{v})=&f(F_0,v_0,0)\\ &+\frac{\partial f}{\partial F_m}\Bigg|_{\substack{F_m=F_0\\v=v_0}}(F_m-F_0)\\ &+\frac{\partial f}{\partial v}\Bigg|_{\substack{F_m=F_0\\v=v_0}}(v-v_0)\\ &+\frac{\partial f}{\partial \dot v}\Bigg|_{\substack{F_m=F_0\\v=v_0}}(\dot{v}-\dot{v_0}) \end{align} Now \begin{align} &\frac{\partial f}{\partial F_m}\Bigg|_{\substack{F_m=F_0\\v=v_0}}=-1\\ &\frac{\partial f}{\partial v}\Bigg|_{\substack{F_m=F_0\\v=v_0}}=C_2+2v_0\\ &\frac{\partial f}{\partial \dot v}\Bigg|_{\substack{F_m=F_0\\v=v_0}}=C_1 \end{align}
I need help with the Taylor expansion.
Q1. In the partial derivatives above, $F_m, v, \dot v$, are treated as constants. Why?
Q2. But aren't they functions of $t$ i.e. $F_m(t),v(t)$ and $\dot v(t)$?
Q3. Isn't the expiclit form of $(1)$ actually $f(F_m(t),v(t),\dot{v}(t))=C_1\dot v(t)+ C_2v(t)+v(t)^2-F_m(t)?$ If so, why no chain rule in the derivatives?
I mean, say we have $g(x(t),y(t))$, then the chain rule gives $\frac{d}{dt}(g(x(t),y(t)))=x'(t)\frac{\partial g}{\partial x}(x(t),y(t)) + y'(t)\frac{\partial g}{\partial y}(x(t),y(t))$
Normally a Taylor series of a function in one dimension, $f(x)$, at the point $x=x^*$ can be written down as
$$ f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(x^*)}{k!}\,(x - x^*)^k. \tag{1} $$
So only the first two terms give
$$ f(x) = f(x^*) + f'(x^*)\,(x - x^*) + O\left((x - x^*)^2\right). \tag{2} $$
When omitting $O\left((x - x^*)^2\right)$, then the right hand side of $(2)$ is the the linearisation of $f(x)$ at $x = x^*$. It can be noted that in $(2)$ $f'(x^*)$ is just a shorter notation for
$$ f'(x^*) = \left.\frac{\partial f(x)}{\partial x}\right|_{x = x^*}. \tag{3} $$
This Taylor series can be generalized to higher dimensions by letting $x$ be a vector of inputs $x = \begin{bmatrix}x_1 & x_2 & \cdots & x_n\end{bmatrix}^\top$ to the function $f(x)$. Now $(3)$ still holds, where the partial derivative with respect to the vector $x$ is defined as follows
$$ \frac{\partial f(x)}{\partial x} = \begin{bmatrix} \frac{\partial f(x)}{\partial x_1} & \frac{\partial f(x)}{\partial x_2} & \cdots & \frac{\partial f(x)}{\partial x_n} \end{bmatrix}. \tag{4} $$
And if $f(x)$ returns a vector as well, then the partial derivative of it with respect to $x$ is also known as its Jacobian.
In your case $x = \begin{bmatrix}F_m & v & \dot{v}\end{bmatrix}^\top$, so the linearisation at $x^* = \begin{bmatrix}F_0 & v_0 & \dot{v}_0\end{bmatrix}^\top$ can be found by using $(4)$ in $(2)$, which gives
$$ f_\text{lin}(x) = f(x^*) + \begin{bmatrix} \frac{\partial f(x)}{\partial F_m} & \frac{\partial f(x)}{\partial v} & \frac{\partial f(x)}{\partial \dot{v}} \end{bmatrix}_{x = x^*} (x - x^*). \tag{5} $$
If you work out the product of the partial derivatives and the difference between the true coordinates and the coordinates at the linearisation point, then you get the same expression as in your question. Just keep in mind that this linearisation is only accurate when difference between $x$ and $x^*$ is small.