Alternative expression of acceleration in vector form

$$\mathbf{a}=\frac{d\mathbf{v}}{dt}=\frac{dx}{dt}\frac{\partial\mathbf{v}}{\partial x}+\frac{dy}{dt}\frac{\partial\mathbf{v}}{\partial y}+\frac{dz}{dt}\frac{\partial\mathbf{v}}{\partial z}=(\mathbf{v}\cdot\nabla)\mathbf{v}$$

I am a bit confused with your notation so I chose my notation.

Assume the components of the position vector $\vec{R}=[x_1,x_2,\ldots,x_{n_R}]^T $are function of the generalized coordinates $q_1,q_2,\ldots,q_{n_Q}$ thus: $x_j=x_j(q_i)$ where $j=1,(1),n_R$ and $i=1,(1),n_Q\quad ,n_Q \le n_R$

We want to obtain the velocity vector $\vec{v}=\frac{d\vec{R}}{dt}$

$$\dot{x}_1=\frac{\partial x_1}{\partial q_1}\,\dot{q}_1+\frac{\partial x_1}{\partial q_2}\,\dot{q}_2+\ldots$$

$$\dot{x}_2=\frac{\partial x_2}{\partial q_1}\,\dot{q}_1+\frac{\partial x_2}{\partial q_2}\,\dot{q}_2+\ldots$$

or $$\dot{x}_j=\sum_i^{nQ}\frac{\partial x_j}{\partial q_i}\,\dot{q}_i$$

or with vector notation (Engineer notation ) :

$$\vec{v}=\vec{\dot R}=\underbrace{\frac{\partial \vec{R}}{\partial \vec{q}}}_{n_R\times n_Q}\,\vec{\dot{q}}$$


$$\vec{R}=\left[ \begin {array}{c} x_{{1}} \left( q_{{1}},q_{{2}} \right) \\ x_{{2}} \left( q_{{1}},q_{{2}} \right) \\ x_{{3}} \left( q_{{1}},q_{{2}} \right) \end {array} \right] =\left[ \begin {array}{c} r\sin \left( q_{{1}} \right) \cos \left( q_{ {2}} \right) \\r\sin \left( q_{{1}} \right) \sin \left( q_{{2}} \right) \\ r\cos \left( q_{{1}} \right) \end {array} \right] $$

$$\vec{q}=\left[ \begin {array}{c} q_{{1}}\\ q_{{2}} \end {array} \right] $$

$$\underbrace{\frac{\partial \vec{R}}{\partial \vec{q}}}_{3\times 2}= \left[ \begin {array}{cc} {\frac {\partial }{\partial q_{{1}}}}x_{{1} } \left( q_{{1}},q_{{2}} \right) &{\frac {\partial }{\partial q_{{2}}} }x_{{1}} \left( q_{{1}},q_{{2}} \right) \\ {\frac { \partial }{\partial q_{{1}}}}x_{{2}} \left( q_{{1}},q_{{2}} \right) &{ \frac {\partial }{\partial q_{{2}}}}x_{{2}} \left( q_{{1}},q_{{2}} \right) \\{\frac {\partial }{\partial q_{{1}}}}x_{ {3}} \left( q_{{1}},q_{{2}} \right) &{\frac {\partial }{\partial q_{{2 }}}}x_{{3}} \left( q_{{1}},q_{{2}} \right) \end {array} \right] $$

Thus: $$\vec{v}=\left[ \begin {array}{cc} r\cos \left( q_{{1}} \right) \cos \left( q_ {{2}} \right) &-r\sin \left( q_{{1}} \right) \sin \left( q_{{2}} \right) \\ r\cos \left( q_{{1}} \right) \sin \left( q_{{2}} \right) &r\sin \left( q_{{1}} \right) \cos \left( q_{{ 2}} \right) \\ -r\sin \left( q_{{1}} \right) &0 \end {array} \right] \,\left[ \begin {array}{c} \dot{q}_{{1}}\\ \dot{q}_{{2}} \end {array} \right] $$


The velocity vector $\vec{v}$ is a function of $\vec{q}$ and $\vec{\dot{q}}$. Your vector v is only a function of $\vec{q}$ this is not the general case