In general, if $\vert \ell m\rangle$ is an eigenstate of $S_z$, then
$$
e^{i\phi S_z/\hbar}\vert \ell m\rangle
=\left(I+i \phi \hbar m/\hbar + \frac{1}{2}(i \phi \hbar m/\hbar)^2+
\ldots\right)\vert \ell m\rangle=e^{im\phi}\vert \ell m\rangle
$$
by definition of the exponential of an operator, and likewise
$$
\langle \ell m\vert e^{i\phi S_z/\hbar}=
\left(e^{i\phi S_z/\hbar}\vert \ell m\rangle\right)^\dagger
=\langle \ell m\vert e^{-im\phi}
$$
so that
$$
e^{i\phi S_z/\hbar}\vert \ell m\rangle\langle \ell m'\vert e^{-i\phi S_z/\hbar} = e^{im\phi} \vert \ell m\rangle\langle \ell m'\vert
e^{im'\phi}
$$
and you can work the rest of the calculation that way.
There is a geometric interpretation to a conjugation like $U \hat A U^\dagger$, where $\hat A$ is an operator: the transformation $U$ is just a change of basis. In your case, $e^{i\phi S_z/\hbar}$ is a change of basis obtained by rotation about $\hat z$ so you would expect under this $S_x$ to go to a linear combination of
$S_x\cos\phi\pm S_y\sin\phi$ and $S_y$ since the $\hat x$ axis rotates to a combination $\hat x\cos\phi\pm \hat y \sin\phi$. The difficulty is with the sign, or alternatively, to understand if $e^{i\phi S_z/\hbar}$ produces a clockwise or anticlockwise rotation.
This is fixed easily enough since, by expanding
\begin{align}
e^{i\phi S_z/\hbar}S_xe^{-i\phi S_z/\hbar}
&=S_x+i\phi [S_z,S_x]+\frac{1}{2!}(i\phi)^2 [S_z,[S_z,S_x]]+\ldots\\
&=S_x+i\phi(iS_y)-\frac{1}{2!}(\phi)^2[S_z,iS_y]+\ldots\\
& =S_x-\phi S_y-\frac{1}{2!}\phi^2 S_x+\ldots
\end{align}
which matches the expansion of $S_x\cos\phi-S_y\sin\phi$.
