Infinitesimal generator of the Brownian motion on a sphere

The generator of Brownian motion on $S^2$ (with the round metric) is $\frac12\Delta$, where $\Delta$ is the Laplacian on $S^2$, in spherical coordinates $$\Delta = \frac{1}{\sin\theta}\partial_{\theta}(\sin\theta\cdot\partial_{\theta})+\frac{1}{\sin^2\theta}\partial_{\varphi}^2.$$ Let $$ X_t=\sin\theta_t\cos\varphi_t,\\ Y_t=\sin\theta_t\sin\varphi_t,\\ Z_t=\cos\theta_t.$$ Now you suggest that if $B_t^{(i)}$, $i=1,2$ are two independent Brownian motions, then $$ d\theta_t = dB_t^{(1)},\\ d\varphi_t = dB_t^{(2)},\tag1$$ defines a Brownian motion on $S^2$. We have $$dZ_t=d\cos\theta_t=-\sin\theta_t\cdot dB_t^{(1)}-\frac12\cos\theta_t\cdot dt.$$ But since $\Delta\cos(\theta)=-2\cos\theta$, we have $$d\cos\theta_t-\frac12\Delta\cos\theta_t=-\frac{3}{2}\cos\theta_t\cdot dt-\sin\theta_t\cdot dB_t^{(1)},$$ Thus $\cos\theta_t-\int_0^t\frac12\Delta\cos\theta_sds$ is not a local martingale. Therefore $\frac12\Delta$ is not the generator of your process (1), and therefore (1) does not define a Brownian motion on $S^2$.

There are many different ways to construct Brownian motion on the sphere. One of them works in Stratonovich form and reads $$ d\mathbf{X}_t = \mathbf{X}_t\otimes d\mathbf{B}_t, \tag2$$ where $\otimes$ denotes a Stratonovich cross product and $\mathbf{B}_t$ is a 3d Brownian motion. In other words, $$ dX_t = Y_t\circ dB^{(3)}_t - Z_t\circ dB^{(2)}_t,\\ dY_t = Z_t\circ dB^{(1)}_t - X_t\circ dB^{(3)}_t, \\ dZ_t = X_t\circ dB^{(2)}_t - Y_t\circ dB^{(1)}_t.$$ First of all we can check that by the Stratonovich chain rule $$ d(X_t^2+Y_t^2+Z_t^2) = 2(X_t\circ dX_t+Y_t\circ dY_t+Z_t\circ dZ_t) = ... = 0,$$ hence $(X_t,Y_t,Z_t)$ is on $S^2$ for all $t\geq 0$ iff $(X_0,Y_0,Z_0)$ is on $S^2$. Then by the Stratonovich chain rule we obtain $$ d\mathbf{X}_t = \frac{\partial \mathbf{x}}{\partial\theta}\circ d\theta_t + \frac{\partial \mathbf{x}}{\partial\varphi}\circ d\varphi_t,$$ By expanding everything and matching with the expressions above, we can solve for $d\theta_t$ and $d\varphi_t$ in terms of the Brownian motions: \begin{align} d\theta_t&=\sin\varphi_t\circ dB_t^{(1)}-\cos\varphi_t\circ dB_t^{(2)},\\ d\varphi_t&=\cot\theta_t\left(\cos\varphi_t\circ dB_t^{(1)}+\sin\varphi_t\circ dB_t^{(2)}\right)-dB_t^{(3)}. \end{align} We get this back into Itô form (please check), which leads to a drift term in $\theta_t$ \begin{align} d\theta_t&=\frac12\cot\theta_t dt+\sin\varphi_t dB_t^{(1)}-\cos\varphi_t dB_t^{(2)},\\ d\varphi_t&=\cot\theta_t\left(\cos\varphi_t dB_t^{(1)}+\sin\varphi_t dB_t^{(2)}\right)-dB_t^{(3)}. \end{align} So now take a $C^2$ function $f(\theta_t,\varphi_t)$ and use Itô's lemma to check that $$df(\theta_t,\varphi_t) = \frac12\Delta f(\theta_t,\varphi_t) dt + ... dB_t^{(1)}+ ... dB_t^{(2)}+ ... dB_t^{(3)}$$ (only the $dt$ term has to be calculated). This shows that the generator of the process (2) is indeed $\frac12\Delta$.

Addendum: By setting \begin{align} dB_t^{\theta}&=\sin\varphi_tdB^{(1)}_t-\cos\varphi_tdB^{(2)}_t,\\ dB^{\phi}_t&=\cos\theta_t(\cos\varphi_tdB^{(1)}_t+\sin\varphi_tdB^{(2)}_t)-\sin\theta_tdB^{(3)}_t \end{align} and checking that these are two independent Brownian motions, one can rewrite the process as \begin{align} d\theta_t&=\frac12\cot\theta_t dt+dB_t^{\theta},\\ d\varphi_t&=\frac{1}{\sin\theta_t}dB^{\phi}_t. \end{align}

There is a third Brownian motion $dB^{N}_t=\mathbf{X}_t\cdot d\mathbf{B}_t$ that is normal to the sphere and therefore gets cancelled out.

Brownian motion on the unit sphere will be given by

$$X_t=(\sin(\theta_t)\cos(\phi_t),\sin(\theta_t)\sin(\phi_t),\cos(\theta_t))$$

where $(\theta_t,\phi_t)$ solves the system of SDEs given by

$$d\theta_t=dB_t^{\theta}+\frac{1}{2}\cot(\theta_t)dt, \qquad d\phi_t=\frac{1}{\sin(\theta_t)}dB_t^{\phi}.$$ Here $B^{\theta}$ and $B^{\phi}$ are independent Brownian motions.

See Why does Brownian motion have drift on Riemannian Manifolds? for the general case.

As an alternative approach for the derivation of the advertised conclusion (in my opinion, this is a more elementary approach). You can take a look at Example 8.5.8 in Oksendal's classical textbook on SDEs, in which the author used a random time change argument to drive the SDE satisfied by a Brownian motion on the unit sphere in $\mathbb{R}^n$ (with $n \geq 3$). This SDE is written in the cartesian coordinates. However, when we consider the special case $n = 3$, a simple Ito-type computations enable us to transform the SDE in cartesian coordinate to the SDE in spherical coordinate $(\theta,\phi)$. I attached the detailed computations below, I hope you can follow and understand.