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 $S_x\cos\phi-S_y\sin\phi$.
