When we take acceleration as $a=dv/dt=v*dT/dt+dv/dt*T$ When $\hat T$ is the unit vector that is tangent to a $\vec r$ vector that describes a particle's movement in space we can write the velocity vector as: $$\vec v=v*\hat T$$
Then if we take the derivative of it with respect to time $t$ to find the acceleration we have:
$$\vec a=\frac{d\vec v}{dt}=v*\frac{d\hat T}{dt}+\frac{dv}{dt}*\hat T$$
We know that if $s$ is the arc length then $dr/ds=1$, so:
$$dr=ds$$
And that:
$$\hat T=\frac{d\vec r}{ds}$$
$$\vec v=\frac{d\vec r}{dt}$$
So we can rewrite the acceleration $a$ the following manner:
$$\vec a=v*\frac{d\hat T}{dt}+\frac{dv}{dt}*\hat T=v*\frac{d\hat T}{ds}*\frac{ds}{dt}+\frac{dv}{dt}*\hat T=\frac{dr}{dt}*\frac{d^2\vec r}{ds^2}*\frac{ds}{dt}+\frac{dv}{dt}*\hat T$$
$$=\frac{d^2\vec r}{dt^2}+\frac{dv}{dt}*\hat T$$
Because $dr=ds$.
Acceleration is $$\vec a=\frac{d^2\vec r}{dt^2}$$
So we end up with:
$$\vec a=\vec a+\frac{dv}{dt}*\hat T$$
Which doesn't make sense since acceleration cannot be equal to itself plus another vector unless that another vector was null but it isn't.
So I must be doing something wrong but I don't know what. I'm not trying to solve any exercise with this, I just want to understand why I'm coming up with this.
 A: To begin with, you make things unnecessarily complicated by introducing
a scalar $r.$ We already have a name, $s,$ for the thing you seem to be trying to call $r.$ So of course we have $dr = ds$ and various other facts, but it
seems simpler just to write
$$ v = \frac{ds}{dt} $$
in the first place, since that's the usual denotation of speed.
The error is the application of a non-existent "chain rule"
for second derivatives.
In general, for variables $x,y,z$ that are functions of each other
such that the total derivatives $\frac{dy}{dx},$
$\frac{d^2z}{dx^2},$ and $\frac{d^2z}{dy^2}$ all exist,
$$
\frac{d^2z}{dy^2} \cdot\left(\frac{dy}{dx}\right)^2 \neq \frac{d^2z}{dx^2}.
$$
Let's try a concrete example with $x, y, z$ all positive:
\begin{align}
y &= x^2  &   z &= y^3 = x^6 \\
\frac{dy}{dx} &= 2x  & 
 \frac{dz}{dy} &= 3y^2 = 3x^4 &  \frac{dz}{dx} &= 6x^5 = 6 y^{5/2} \\
&& \frac{d^2z}{dy^2} &= 6y = 6x^2 &  \frac{d^2z}{dx^2} &= 30x^4 = 30 y^2
\end{align}
\begin{align}
\frac{d^2z}{dy^2} \cdot \left(\frac{dy}{dx}\right)^2
 &= \left(6x^2\right)\left(2x\right)^2 = 24x^4 \\
 24x^4  &\neq 30x^4 \\
\frac{d^2z}{dy^2} \cdot \left(\frac{dy}{dx}\right)^2
 &\neq \frac{d^2z}{dx^2}.
\end{align}
The equations above are all dealing with scalars and you are dealing with both scalars and vectors, but the false "second derivative chain rule"
does not suddenly become true when you throw some vectors into the mix.
You may wonder why the following is not a valid sequence of equations:
\begin{align}
\frac{d^2z}{dy^2} \cdot \left(\frac{dy}{dx}\right)^2
&\stackrel?=
\frac{d}{dy} \left(\frac{dz}{dy}\right) \cdot \left(\frac{dy}{dx}\right)^2
\stackrel?=
\frac{d}{dx} \left(\frac{dz}{dy}\right) \cdot \left(\frac{dy}{dx}\right)\stackrel?= \frac{d^2z}{dx\,dy} \cdot \left(\frac{dy}{dx}\right) \\
&\stackrel?= \frac{d^2z}{dy\,dx} \cdot \left(\frac{dy}{dx}\right)
\stackrel?=
\frac{d}{dy} \left(\frac{dz}{dx}\right) \cdot \left(\frac{dy}{dx}\right)
\stackrel?= \frac{d}{dx} \left(\frac{dz}{dx}\right)  \\
&\stackrel?= \frac{d^2z}{dx^2}.
\end{align}
The problem is here:
$$
\frac{d^2z}{dx\,dy} \stackrel?= \frac{d^2z}{dy\,dx}.
$$
To assert that the two sides of this formula are always equal
is tantamount to an assertion that the total derivative operators 
$\frac{d}{dx}$ and $\frac{d}{dy}$ commute.
They do not:
\begin{align}
\frac{d}{dx} \left(\frac{dz}{dy}\right) 
  &= \frac{d}{dx} \left(3x^4\right) = 12x^3 \\
\frac{d}{dy} \left(\frac{dz}{dx}\right) 
  &= \frac{d}{dy} \left(6 y^{5/2}\right) = 15 y^{3/2} = 15x^3.
\end{align}
It would be different if we were dealing with the partial differentiation
operators $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$
instead of total differentiation; those operators do commute.
But many things would be different if we were dealing with partial differentiation.
