Suppose there is a vector $vec v$ which is a function of time, then will $dfrac{d}{dt}|vec v|$ be a vector quantity or a scalar quantity?

$a_t=d|v_{t}|/dt$

This only gives the magnitude of the tangential acceleration, overall tangential acceleration is a vector quantity.

The careful mathematics goes like this:

The rate of change of the speed of the particle is given by $$ frac{dv}{dt} = frac{d}{dt}sqrt{vec{v}cdotvec{v}}. $$ Using the chain and product rules of differentiation, we get $$ frac{dv}{dt} = frac{1}{2sqrt{vec{v}cdotvec{v}}}frac{d}{dt}vec{v}cdotvec{v} = frac{1}{2v}left(frac{dvec{v}}{dt}cdotvec{v}+vec{v}cdotfrac{dvec{v}}{dt}right) = frac{1}{2v}left(vec{a}cdotvec{v}+vec{v}cdotvec{a}right) = frac{vec{a}cdotvec{v}}{v} = vec{a}cdothat{v}, $$ where $hat{v}$ is the unit vector in the direction of $vec{v}$, so $hat{v}$ it is the direction of $vec{v}$. From this, we can see that since we are dotting the acceleration into the velocity, we get the component of $vec{a}$ along $hat{v}$ that leads to changes in speed. This component is what you would call $a_t = vec{a}cdothat{v}$, and because of the dot-product, it is manifestly not a vector quantity.

Next, we look at how the direction of $vec{v}$ is changing. Since the direction of $vec{v}$ is just $hat{v}$, we want to compute the derivative of $hat{v}$: $$ frac{dhat{v}}{dt} = frac{d}{dt}frac{vec{v}}{v} = frac{1}{v}frac{dvec{v}}{dt} - vec{v}frac{1}{v^2}frac{dv}{dt}, $$ where we again used the product rule (first) and then the chain rule. We rearrange this equation carefully and substitute in for $dv/dt$ from our previous calculation, resulting in $$ frac{dhat{v}}{dt} = frac{1}{v}left(vec{a} - (vec{a}cdothat{v})hat{v}right). $$ The quantity in parentheses is exactly the component of $vec{a}$ perpendicular to the velocity. (You can check orthogonality by taking the dot product of this vector with $vec{v}$ and finding that it's zero.) The change in direction $dhat{v}/dt$ therefore depends only on this perpendicular component, which we might call $a_r = vec{a} - (vec{a}cdothat{v})hat{v}$.