Given any vector ##\vec A## and any vector ##\vec V##,
one can write ##\vec A## as the sum of two vectors:
one parallel to ##\vec V## and the other (the rest of ##\vec A##) perpendicular to ##\vec V##.
\begin{align*}
\vec A
&=\vec A_{||V} + (\vec A - \vec A_{||V})
\end{align*}Now,
\begin{align*}
\vec A_{||V}
&= (\vec A \cdot \hat V) \hat V\\
&= (\vec A \cdot \frac{\vec V}{V}) \frac{\vec V}{V}
\end{align*}
[As a special case, let \hat V = \hat x.]
(Check that \vec A_{||V}\cdot \hat V=(\vec A \cdot \hat V)
and that (\vec A - \vec A_{||V})\cdot \hat V=0.)
For a trajectory, \hat V is tangent to the trajectory and (for a planar curve)
\hat V_{\bot} is normal to the trajectory.
So, if \vec A is the acceleration vector (\vec A =\frac{d}{dt}\vec V),
then \vec A_{||V} is the tangential-part of the acceleration vector (responsible for changing the magnitude of \vec V [i.e. speeding up or slowing down]),
and the other part is the normal-part of the acceleration vector (responsible for changing the direction of \vec V [turning the velocity vector] ).
Next, one has to express these in terms of the angular pseudovectors \vec \alpha and \vec \omega [probably should use a different "arrowhead"] (which when crossed with ordinary [polar] vectors result in ordinary [polar] vectors).
You'll probably [implicitly] use the "BAC-CAB" rule
https://en.wikipedia.org/wiki/Triple_product#Vector_triple_product
https://mathworld.wolfram.com/BAC-CABIdentity.html
