The equation you have given is not quite right, but you are on the right track. This is fairly simple if you remember that the characteristic function is just an expected value of a complex exponential. As such, it obeys all the normal rules for the expected value of a function of variable. In general, the characteristic function of $X+Y$ can be written as:
$$\begin{equation} \begin{aligned}
\phi_{X+Y}(t)
&\equiv \mathbb{E}(e^{it(X+Y)}) \\[6pt]
&= \mathbb{E}(e^{itX} e^{itY}) \\[6pt]
&= \mathbb{E}(e^{itX}) \cdot \mathbb{E}(e^{itY}) + \mathbb{Cov}(e^{itX}, e^{itY}) \\[6pt]
&= \phi_{X}(t) \cdot \phi_{Y}(t) + \mathbb{Cov}(e^{itX}, e^{itY}). \\[6pt]
\end{aligned} \end{equation}$$
In the case where $X$ and $Y$ are independent, we have $e^{itX} \bot \ e^{itY}$ which gives $\mathbb{Cov}(e^{itX}, e^{itY})=0$, so we get the simpler rule $\phi_{X+Y}(t) = \phi_{X}(t) \cdot \phi_{Y}(t)$. Alternatively, taking $\phi_{Y|X}(t|x) \equiv \mathbb{E}(e^{itY} |X=x)$ you can use the law of iterated expectation to get:
$$\begin{equation} \begin{aligned}
\phi_{X+Y}(t)
&\equiv \mathbb{E}(e^{it(X+Y)}) \\[6pt]
&= \mathbb{E}(e^{itX} e^{itY}) \\[6pt]
&= \mathbb{E}(\mathbb{E}(e^{itX} e^{itY}|X)) \\[6pt]
&= \mathbb{E}(e^{itX} \cdot \mathbb{E}(e^{itY} |X)) \\[6pt]
&= \mathbb{E}(e^{itX} \cdot \phi_{Y|X}(t|X)) \\[6pt]
\end{aligned} \end{equation}$$
Note that the conditional characteristic function is not generally separable from the expectation over $X$ in this case (which is why your equation is wrong).