I'm having trouble understanding the intuition behind why $a(x) = v\frac{\mathrm{d}v}{\mathrm{d}x}$ I was shown
\begin{align}
a(x) &= \frac{\mathrm{d}v}{\mathrm{d}t}\\
&= \frac{\mathrm{d}v}{\mathrm{d}x}\underbrace{\frac{\mathrm{d}x}{\mathrm{d}t}}_{v}\\
&= v\frac{\mathrm{d}v}{\mathrm{d}x}
\end{align}
However, this feels somewhat unintuitive, and somewhat questionable mathematics-wise. Perhaps it's the best way to explain it, but I was hoping for a more intuitive understanding of this formula.
 A: For a simple function like $x=t^2$ you can show that the chain rule works.
$x=t^2 \to \dot x = v = 2t \to \ddot x = \dot v = a = 2$
$t=x^{1/2} \to v=2x^{1/2} \to \frac{dv}{dx} = x^{-1/2}= 1/t$
$\frac {dv}{dx} \cdot v  = 1/t \cdot 2t = 2 = a$

Looking at the slopes of the graphs you can imagine that as time progresses the increasing gradient of the one on the left multiplied by the decreasing gradient of the one in the middle could produce a constant value for the graph on the right.
A: Writing for a general case, $v$ can be an explicit function of both $t$ and $x$ (for 1D motion along $x$).
$\therefore$ \begin{equation}
dv=\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial t}dt \Rightarrow \frac{dv}{dt}=\frac{\partial v}{\partial x}\frac{dx}{dt}+\frac{\partial v}{\partial t}=a \quad ...(\star)
\end{equation}
if $v$ is an explicit function of $x$ only($v=v(x))$, $\frac{\partial v}{\partial t}=0 $ and $\frac{\partial v}{\partial x}=\frac{dv}{dx}$
$\therefore$ ($\star$) becomes: \begin{equation}
a=\frac{dv}{dx}\frac{dx}{dt}=\left(\frac{dv}{dx}\right)v
\end{equation}
A: This intuition is for one dimensional motion: Consider a particle moving in a line. At any time $t$, its velocity is $v(t)$, its acceleration is $a(t)$. The change in velocity after a small time $dt$ will be $a(t)dt$. The change in position will be $dx=v(t)dt$
The identity you wrote is expressing the fact that $vdv=adx$. This can be verified: $vdv=v(adt)=a(vdt)=adx$
Keep in mind that you do not need $v$ to be a function of $x$ for this identity to work. It's only saying that, at any time $t$, $a(t)$ multiplied by $dx$ will be equal to $v(t)$ multiplied by $dv$. $dx$ and $dv$ are the changes in position and velocity after an infinitesimally small time $dt$.
For three dimensional motion, the identity is $\vec{a}\cdot \vec{dx}=\vec{v}\cdot \vec{dv}$.
A: the kinematic equations are
$$x(t)=f(t)\quad\Rightarrow\\
v(t)=\frac{dx(t)}{dt}=\frac{df(t)}{dt}\\
a(t)=\frac{dv(t)}{dt}= \frac{d^2f(t)}{dt^2}$$
now if you want to obtain the acceleration $~a=a(x)~$ first you eliminate the parameter $~t~$ with the equation $~x=f(t)\quad\Rightarrow~t=g(x)~$ hence
$$a(x)=\frac{d^2f(t)}{dt^2}\bigg|_{t=g(x)}$$
but also with:
$$t=g(x)\quad,dt=\frac{\partial g(x)}{\partial x}\,dx\quad\Rightarrow\\
v(x)=\frac{dx}{dt}=\frac {1}{\frac{\partial g(x)}{\partial x}}\\
a(x)=\frac{dv}{dt}=\frac{dv(x)}{dx}\,\frac{dx}{dt}=\frac{dv(x)}{dx}\,v(x)$$
Example:
$$x=f(t)=c\,t^2\quad\Rightarrow\\
t=\frac {\sqrt{c\,x}}{c}=g(x)\quad,g'(x)=\frac{1}{2\sqrt{c\,x}}\\
v(x)=2\,\sqrt{c\,x}\quad,v'(x)=\frac{c}{\sqrt{c\,x}}\\
a(x)=\frac{c}{\sqrt{c\,x}}\,2\,\sqrt{c\,x}=2\,c $$
or
$$a(x)=\frac{d^2f(t)}{dt^2}\bigg|_{t=g(x)}=2\,c $$

also
$$a(x)=-{\frac {{\frac {d^{2}}{d{x}^{2}}}g \left( x \right) }{ \left( {\frac 
{d}{dx}}g \left( x \right)  \right) ^{3}}}
$$
A: It's just the chain rule. Alternatively, the change of kinetic energy must be the work done:
$$d( m v^2/2) = F \,dx \quad \Rightarrow \quad m \,v\, dv \,=\, m  a\, dx
\quad \Rightarrow \quad v \frac{dv}{dx} =  a$$
Note: the general mathematical fact $d(q^2)=2q\,dq$ has been used to treat the differential of $v^2$.
A: If you turn the expression into
$$ a\, {\rm d}x = v\, {\rm d}v \tag{1}$$
and integrate both sides, you get
$$ \int  a\, {\rm d}x = \tfrac{1}{2} v^2 \tag{2}$$
Now multiply both sides with $m$ and note that $F= m a$ to get
$$ \int F \,{\rm d}x = \tfrac{1}{2}m v^2 \tag{3}$$
or commonly known as the work-energy theorem. Total work done equals to kinetic energy (or change in KE).
Now that you understand (3) work backward towards (1) to see that the differential work done per unit mass equals the differential change in kinetic energy.
A: There are many good answers here which address the mathematics. I will attempt to give you something to imagine.
Rather than imagine a single particle, imagine many particles moving at once, perhaps even in 1 dimension. You can imagine a conveyer belt carrying particles: each particle started at the beginning, but they started at the beginning at different times, so that we can imagine them simultaneously at the current time.
(Later, to return to a single particle, then this is flipped: if you imagine many particles starting at the beginning at the same time, they stack and act as 1 single particle, but then you may have difficulty imagining the particle simultaneously at different times.)
Let each particle be so small, and let there be so many particles, that for a length $dx$ you can find another particle. That is, maybe particle A is at position x, and particle B is at position (x+dx). Certainly, they have different positions, and perhaps they have different velocities as well. If they have different velocities, it will be because of acceleration, since particle B is simply living the same life track (trajectory) as particle A, but started earlier and hence is further along.
Now let's figure out how an acceleration might appear.
Starting from the spatial difference, if we know the velocity, we can figure out how much of a head start particle B had. B is further along in the race, so it started earlier, but how much earlier? Well, for A to reach the position where B is now, A must move $dx$ with velocity $v$, so A needs time $dt = dx/v$ to reach B's place in life.
If we know the difference in velocity between A and B, $dv$, then that means we know $dv/dx$: a difference in velocity given a difference in position, since the two particles are at different positions: $dv = (dv/dx) * dx$.
We know this difference in velocity is due to an acceleration acting over the head-start time: $dv = a * dt = a * dx / v$. This is another way of thinking of the velocity difference between A and B, but we know two ways of thinking of one thing must be equal, hence $a dx/v = dv = (dv/dx) * dx$, or $a = v (dv/dx)$.
Returning to a single particle, replace the comparison between A and B with Present A (at time t and position x and velocity v) vs Future A (at time t+dt and position x+dx and velocity v+dv).
Lastly, I briefly mention the material derivative as an example of this principle, in case somehow fluid dynamics helps you gain intuition about it.
