(1) The infinitesimal generator is an operator defined on a subspace of the space $C_0$ of continuous functions that vanish at infinity. More precisely, the subspace on which it is defined is
$$D(A) = \left \{f \in C_0: \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(X_t))-f(x)}{t} \text{ exists} \right \}$$
and then $A:D(A) \to C_0$. This means that if $f \in D(A)$ then $Af$ is a function that vanishes at infinity. $A$ is the generator of the process $X$ but this does not mean that for $f \in D(A)$, $Af(X) = AX$. Indeed, you have no reason to believe that $X \in D(A)$ in the first place since in many situations $A$ will turn out to be a differential operator and $X$ need not have differentiable paths.
Finally, if $X$ is your process then $f(X)$ is the process that at a given time $t$ and for a given $\omega$ in your probability space is defined by $f(X)_t(\omega) = f(X_t(\omega))$.
(2) $P_t f(x) = \mathbb{E}^x[f(X_t)]$. This is simply the expected value of $f(X_t)$ is your process is started at $x$ at time $0$.
(3) $f(x)$ is deterministic and so $E^x[f(x)] = f(x) E^x[1] = f(x)$.
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.
Best Answer
Starting from $$dY_1=-\frac12 Y_1 dt-Y_2 dB_t, \\ dY_2=-\frac12 Y_2 dt+Y_1 dB_t, $$
and using the general formula for the generator of a diffusion process $dX_t=f(X_t)dt+g(X_t)dB_t$, which reads $$\mathcal{A}=\sum_{i=1}^n f_i(x)\partial_{x_i}+\frac{1}{2}\sum_{i=1}^n\sum_{j=1}^m g_{ij}(x)\partial_{x_i}\partial_{x_j},$$ we have with $n=2$ and $m=1$ $$\mathcal{A}=\frac12\left(-y_1\partial_{y_1}-y_2\partial_{y_2}+y_2^2\partial_{y_1}^2+y_1^2\partial_{y_2}^2\right)=\frac12\partial_{\theta}^ 2.$$ Which is one half times the Laplacian on $S_1$.
In fact, one of the definitions of Brownian motion on a Riemannian manifold $(M,g)$ is that its generator is $\frac12\Delta_g$, where $\Delta_g$ is the Laplacian of $g$. Thus in order to find the generator of Brownian motion on $S_2$ in Cartesian coordinates, take the Laplacian on $S_2$ and transform it to Cartesian coordinates (a slightly tedious calculation).