An interresting "method" that allows you to know the acceleration vector with respect to any coordinate system is just a matter of recognize some key formulas.
1) Given the metric of a particlar line elemente of a particular coordinate system. Find the metric tensor:
$$ds^{2} = g_{\gamma \beta} dx^{\gamma}dx^{\beta}$$
2) Given the "definition" of Christoffel symbols. Compute all the Christoffel symbols:
$$ \displaystyle \Gamma ^{\alpha} _{\mu \nu} = g^{\delta \alpha} \Big\{ g_{\mu \delta,\nu} + g_{\nu \delta,\mu} - g_{\mu \nu,\delta} \Big\} $$
3) Calculate the "generalized force" components:
$$ \vec{F} = m\vec{a} = m \Big(a^{\alpha}\frac{\partial}{\partial x^{\alpha}} \Big) = \Big(m a^{\alpha} \Big) \frac{\partial}{\partial x^{\alpha}} = f^{\alpha} \frac{\partial}{\partial x^{\alpha}} $$
$$ f^{\alpha} = \Big(m a^{\alpha}\Big) = m \Big( \frac{d^{2}x^{\alpha}}{dt^{2}}+\displaystyle \Gamma ^{\alpha} _{\mu \nu} \frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt} \Big) $$
4) Calculate The "physical components" of the generalized force (contravariant) vector and unitary basis vectors for spherical coordinate system:
According to Sochi,pages (19-20) [https://arxiv.org/pdf/1610.04347.pdf] :
For a contravariant vector we have:
$$ \vec{A} = A^{\mu}\frac{\partial}{\partial x^{\mu}}$$
Now, the Physical Representation of basis (covariant) vectors (unitary basis vectos) are:
$$\hat{e}_{\mu}= \frac{\partial}{\partial x^{\mu}} \frac{1}{\sqrt{g_{\mu \mu}}} \equiv \frac{\partial}{\partial x^{\mu}} \frac{1}{h_{\mu}}$$ (No Sum in $\mu$)
And the Physical Representation for (Contravariant) componentes are:
$$A^{\mu}_{physical} = \sqrt{g_{\mu \mu}} A^{\mu} \equiv h_{\mu}A^{\mu}$$
Then, for the force we have:
$$ \vec{F}_{physical} = m \vec{a}_{physical} = \Big[\sqrt{g_{\alpha \alpha}}f^{\alpha}\Big] \Big[ \frac{1}{\sqrt{g_{\alpha \alpha}}} \frac{\partial}{\partial x^{\alpha}} \Big]$$
5) Setting $m=1$, then you have the accerelation (physical) vector :
$$ \vec{F}_{physical} = 1 \vec{a}_{physical}$$
$$ \vec{a}_{physical} = \Big[\sqrt{g_{\alpha \alpha}}a^{\alpha}\Big] \Big[ \frac{1}{\sqrt{g_{\alpha \alpha}}} \frac{\partial}{\partial x^{\alpha}} \Big]$$
For instance:
1) We know the metric tensor from line element of spherical coordinates:
$$ds^{2} = dr^{2} + r^{2}d\theta^{2} + r^{2}sin^{2}(\theta) d\phi^{2} \implies $$
$$ g^{spherical}_{\gamma \beta} = Diag (1, r^{2}, r^{2}sin^{2}(\theta))$$
2) The non zero Christoffel symbols are:
$$\Gamma ^{r}_{\theta \theta} = -r \\ \Gamma ^{r}_{\phi \phi} = -rsin^{2}(\theta) \\ \Gamma ^{\theta}_{\phi \phi} = -sin(\theta) cos(\theta) \\ \Gamma ^{\theta}_{r\theta} = \Gamma ^{\theta}_{\theta r} = \frac{1}{r}\\ \Gamma ^{\phi}_{r\phi} = \Gamma ^{\phi}_{\phi r} = \frac{1}{r} \\ \Gamma ^{\phi}_{\theta \phi} = \Gamma ^{\phi}_{\phi \theta} = cotg(\theta)$$
3) The generalized force components are:
$$ f^{r} = m\Big( \ddot{r} - r\dot{\theta}^2 - rsin^{2}(\theta) \dot{\phi}^{2} \Big)$$
$$f^{\theta} = m\Big(\ddot{\theta} +2\frac{1}{r}\dot{r}\dot{\theta} - sin(\theta)\cos(\theta)\ddot{\phi}^{2} \Big)$$
$$f^{\phi} = m\Big(\ddot{\phi} +2\frac{1}{r}\dot{r}\dot{\theta} + 2cotg(\theta) \dot{\phi} \dot{\theta} \Big)$$
Then,the force vector is:
$$\vec{F} = f^{r}\frac{\partial}{\partial x^{r}} + f^{\theta}\frac{\partial}{\partial x^{\theta}}+f^{\phi}\frac{\partial}{\partial x^{\phi}} \implies $$
$$\vec{F} =\Big[ m\Big( \ddot{r} - r\dot{\theta}^2 - rsin^{2}(\theta) \dot{\phi}^{2} \Big) \Big] \frac{\partial}{\partial x^{r}}\\+ \Big[ m\Big(\ddot{\theta} +2\frac{1}{r}\dot{r}\dot{\theta} - sin(\theta)\cos(\theta)\ddot{\phi}^{2} \Big) \Big]\frac{\partial}{\partial x^{\theta}}\\+\Big[ m\Big(\ddot{\phi} +2\frac{1}{r}\dot{r}\dot{\theta} + 2cotg(\theta) \dot{\phi} \dot{\theta} \Big) \Big]\frac{\partial}{\partial x^{\phi}} $$
4) The physical components and vector are then:
$$\vec{F}_{physical} = \Big[\sqrt{g_{\alpha \alpha}}f^{\alpha}\Big] \Big[ \frac{1}{\sqrt{g_{\alpha \alpha}}} \frac{\partial}{\partial x^{\alpha}} \Big] = \sqrt{g_{rr}}f^{r} \frac{1}{\sqrt{g_{rr}}} \frac{\partial}{\partial x^{r}} + \sqrt{g_{\theta \theta}}f^{\theta} \frac{1}{\sqrt{g_{\theta \theta}}} \frac{\partial}{\partial x^{\theta}}+ \sqrt{g_{\phi \phi}}f^{\phi} \frac{1}{\sqrt{g_{\phi \phi}}} \frac{\partial}{\partial x^{\phi}} \implies $$
$$\vec{F}_{physical}= \Big[ m\Big( \sqrt{g_{rr}}\ddot{r} -\sqrt{g_{rr}} r\dot{\theta}^2 - \sqrt{g_{rr}}rsin^{2}(\theta) \dot{\phi}^{2} \Big) \Big]\frac{1}{\sqrt{g_{rr}}} \frac{\partial}{\partial x^{r}}\\+ \Big[ m\Big( \sqrt{g_{\theta \theta}}\ddot{\theta} +\sqrt{g_{\theta \theta}}2\frac{1}{r}\dot{r}\dot{\theta} -\sqrt{g_{\theta \theta}} sin(\theta)\cos(\theta)\ddot{\phi}^{2} \Big) \Big] \frac{1}{\sqrt{g_{\theta \theta}}} \frac{\partial}{\partial x^{\theta}}\\+\Big[ m\Big(\sqrt{g_{\phi \phi}}\ddot{\phi} +\sqrt{g_{\phi \phi}}2\frac{1}{r}\dot{r}\dot{\theta} +\sqrt{g_{\phi \phi}} 2cotg(\theta) \dot{\phi} \dot{\theta} \Big) \Big] \frac{1}{\sqrt{g_{\phi \phi}}}\frac{\partial}{\partial x^{\phi}} \implies$$
$$\vec{F}_{physical}=\Big[ m\Big( 1 \ddot{r} -1 r\dot{\theta}^2 - 1 rsin^{2}(\theta) \dot{\phi}^{2} \Big) \Big] \frac{1}{1}\frac{\partial}{\partial x^{r}}\\+ \Big[ m\Big( r \ddot{\theta} +r2\frac{1}{r}\dot{r}\dot{\theta} -r sin(\theta)\cos(\theta)\ddot{\phi}^{2} \Big) \Big] \frac{1}{r}\frac{\partial}{\partial x^{\theta}}\\+\Big[ m\Big( \sqrt{r^{2}sin^{2}(\theta)}\ddot{\phi} +\sqrt{r^{2}sin^{2}(\theta)}2\frac{1}{r}\dot{r}\dot{\theta} +\sqrt{r^{2}sin^{2}(\theta)} 2cotg(\theta) \dot{\phi} \dot{\theta} \Big) \Big] \frac{1}{rsin(\theta)}\frac{\partial}{\partial x^{\phi}} \implies$$
5) Setting $m=1$, we have the well known acceleration vector in Spherical Coordinares:
$$ \vec{a}_{physical} =\Big( \ddot{r} -r\dot{\theta}^2 - rsin^{2}(\theta) \dot{\phi}^{2} \Big)\frac{1}{1}\frac{\partial}{\partial x^{r}}\\+\Big( r \ddot{\theta} +2\dot{r}\dot{\theta} -r sin(\theta)\cos(\theta)\ddot{\phi}^{2} \Big) \frac{1}{r}\frac{\partial}{\partial x^{\theta}}\\+\Big( rsin(\theta) + 2sin(\theta)\dot{r}\dot{\phi}+2rcos(\theta)\dot{\phi}\dot{\theta} \Big)\frac{1}{rsin(\theta)}\frac{\partial}{\partial x^{\phi}} \implies$$
$$ \vec{a}_{physical} =\Big( \ddot{r} -r\dot{\theta}^2 - rsin^{2}(\theta) \dot{\phi}^{2} \Big)\hat{e}_{r}\\+\Big( r \ddot{\theta} +2\dot{r}\dot{\theta} -r sin(\theta)\cos(\theta)\ddot{\phi}^{2} \Big) \hat{e}_{\theta}\\+\Big( rsin(\theta) + 2sin(\theta)\dot{r}\dot{\phi}+2rcos(\theta)\dot{\phi}\dot{\theta} \Big)\hat{e}_{\phi}$$
Now, my question is: how can I compute the velocity and position vectors?
The reason for these awful calculations is simple, you just need to know a few formulas and how to calculate partial derivatives properly. Then you earn a powerful method to compute the equations of motion in any coordinate system (also in a flat and curved space).