Assume deterministic and constant interest rates.
For an investor in the foreign economy i.e. a market participant that can only trade assets delivering a payout in the foreign currency, let us define
$$ \tilde{X}_t = \tilde{X}_0 \exp \left(\left(r_f-r_d-\frac{\sigma_\tilde{X}^2}{2}\right)+\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f} \right) $$
$$ Y_t =Y_0\exp \left(\left(r_f-\frac{\sigma_Y^2}{2}\right)t+\sigma_Y W_t^{Y,\mathbb{Q}^f} \right) $$
where
- $\mathbb{Q}^f$ figures the foreign risk-neutral measure (risk-free MMA $B^f_t = \exp(r_f\ t)$ is the numéraire).
- $\tilde{X}_t$ representing the instantaneous DOM/FOR exchange rate. $\tilde{X}_t = \text{x}$ means that, at time $t$, 1 unit of domestic currency equals $\text{x}$ units of foreign currency.
- $Y_t$ an equity underlying denominated in the foreign currency.
Let's further assume that the 2 Brownian motions $W_t^{\tilde{X},\mathbb{Q}^f}$ and $W_t^{Y,\mathbb{Q}^f}$ are correlated
$$ d\langle W^{\tilde{X},\mathbb{Q}^f}, W^{Y,\mathbb{Q}^f} \rangle_t = \rho dt $$
Notice how I have used $\tilde{X}_t$ (DOM/FOR) and not $X_t$ (FOR/DOM) as you propose, because in the foreign economy, the only tradable assets are: $Y_t$, $B^f_t$ and $B^d_t \tilde{X}_t$ as hinted above (and these should be all $\mathbb{Q}^f$-martingales when expressed under the numéraire $B_t^f$). We do have the relationship, $\tilde{X}_t = 1/X_t $.
Thanks to the fundamental theorem of asset pricing, for any tradable asset $V_t$ denominated in the foreign currency, we have that, under the foreign risk-neutral measure $\mathbb{Q}^f$
$$ \frac{V_t}{B^f_t} \text{ is a } \mathbb{Q}^f \text{- martingale} \iff \frac{V_0}{B^f_0} = E^{\mathbb{Q}^f}_0 \left[ \frac{V_t}{B^f_t} \right] $$
Under the domestic risk-neutral measure $\mathbb{Q}^d$ (risk-free MMA $B^d_t = \exp(r_d\ t)$ is the numéraire)
$$ \frac{V_t/\tilde{X}_t}{B^d_t} \text{ is a } \mathbb{Q}^d \text{- martingale} \iff \frac{V_0/\tilde{X}_0}{B^d_0 } = E^{\mathbb{Q}^d}_0 \left[ \frac{V_t /\tilde{X}_t^d}{B^d_t} \right] $$
in words, the foreign asset value converted to domestic currency units is a martingale under the domestic risk-neutral measure.
From the above, we see that the Radon-Nikodym derivative writes
$$ \left. \frac{d\mathbb{Q}^d}{d\mathbb{Q}^f} \right\vert_{\mathcal{F}_0} = \frac{B_0^f B_t^d \tilde{X}_t}{B_t^f B_0^d \tilde{X}_0} $$
yet because
\begin{align}
\tilde{X}_t &= \tilde{X}_0\exp \left(\left(r_f-r_d-\frac{1}{2}\sigma_\tilde{X}^2\right)t+\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f} \right) \\
&= \tilde{X}_0 \frac{B^f_t}{B^f_0}\frac{B^d_0}{B^d_t}\exp \left(-\frac{1}{2}\sigma_\tilde{X}^2t+\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f} \right)
\end{align}
this deriative also writes
\begin{align}
\left. \frac{d\mathbb{Q}^d}{d\mathbb{Q}^f} \right\vert_{\mathcal{F}_0} &= \exp\left(\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f}-\frac{1}{2}\sigma_\tilde{X}^2t\right) \\
&= \mathcal{E}\left(\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f} \right)
\end{align}
which is indeed a well-behaved Doléans-Dade exponential where we've used the notation
$$\mathcal{E}(M_t) = \exp \left( M_t - \frac{1}{2}\langle M \rangle_t \right)$$
to denote the stochastic exponential.
Hence Girsanov theorem can be applied to transform Brownian motions under $\mathbb{Q}^f$ as Brownian motions under $\mathbb{Q}^d$. How does it work?
Girsanov Theorem (non rigourous version) - Let $W_t^{\mathbb{Q^f}}$ represent a standard Brownian motion under $\mathbb{Q^f}$ and assume the Radon-Nikodym derivative can be written as:
$$ \left. \frac{d\mathbb{Q}^d}{d\mathbb{Q}^f} \right\vert_{\mathcal{F}_0} = \mathcal{E}(L_t) $$
In that case, the process $W_t^{\mathbb{Q^d}}$ defined as
$$ W_t^{\mathbb{Q^d}} = W_t^{\mathbb{Q^f}} - \langle W^{\mathbb{Q^f}}, L \rangle_t $$
is a standard a Brownian motion under $\mathbb{Q^d}$.
In our particular example, we see that
$$L_t := \sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f} $$
Applying Girsanov theorem then allows us to write
\begin{align}
W_t^{\tilde{X},\mathbb{Q}^d} &= W_t^{\tilde{X},\mathbb{Q}^f} - \langle W^{\tilde{X},\mathbb{Q}^f}, \sigma_\tilde{X} W^{\tilde{X},\mathbb{Q}^f} \rangle_t \\
&= W_t^{\tilde{X},\mathbb{Q}^f} - \sigma_\tilde{X}t \\
W_t^{Y,\mathbb{Q}^d} &= W_t^{Y,\mathbb{Q}^f} - \langle W^{Y,\mathbb{Q}^f}, \sigma_\tilde{X} W^{\tilde{X},\mathbb{Q}^f} \rangle_t \\
&= W_t^{Y,\mathbb{Q}^f} - \rho \sigma_\tilde{X} t
\end{align}
meaning that, to move from $\mathbb{Q}^f$ to $\mathbb{Q}^d$ one can just replace
\begin{align}
W_t^{\tilde{X},\mathbb{Q}^f} = W_t^{\tilde{X},\mathbb{Q}^d} + \sigma_\tilde{X} t \\
W_t^{Y,\mathbb{Q}^f} = W_t^{Y,\mathbb{Q}^d} + \rho \sigma_\tilde{X} t \\
\end{align}
in the expressions for $\tilde{X}_t$ and $Y_t$ to obtain:
\begin{align}
\tilde{X}_t = \tilde{X}_0 \exp \left(\left(r_f - r_d + \frac{\sigma_\tilde{X}^2}{2}\right) t + \sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^d} \right) \\
Y_t = Y_0 \exp \left(\left(r_f + \rho \sigma_\tilde{X} \sigma_Y - \frac{\sigma_Y^2}{2}\right) t + \sigma_Y W_t^{Y,\mathbb{Q}^d} \right)
\end{align}
Now assume we want to compute the expectation of $Y_tX_t = Y_t/\tilde{X}_t$ under $\mathbb{Q}^d$. The random variable $Y_t/\tilde{X}_t$ being lognormally distributed (ratio of two lognormals) with mean
$$ \mu = \ln(Y_0/\tilde{X}_0) + \left(r_d - \frac{\sigma^2_X - 2\rho\sigma_\tilde{X}\sigma_Y + \sigma_Y^2}{2}\right)t $$
and variance
$$ \sigma^2 = \left(\sigma_\tilde{X}^2 - 2 \rho \sigma_\tilde{X} \sigma_Y + \sigma_Y^2 \right)t $$
applying the usual formula gives
\begin{align}
E^{\mathbb{Q}^d}[Y_t/\tilde{X}_t] &= \exp \left(\mu+\frac{\sigma^2}{2} \right) \\
&= Y_0/\tilde{X}_0 \exp \left(r_d t \right) \\
&= Y_0/\tilde{X}_0 B_t^d
\end{align}
hence
$$ E^{\mathbb{Q}^d} \left[ \frac{Y_t/\tilde{X}_t}{B_t^d} \right] = \frac{Y_0/\tilde{X}_0}{B_0^d} $$
as it should since we already knew that
$$ \frac{Y_t/\tilde{X}_t}{B^d_t} \text{ was a } \mathbb{Q}^d \text{- martingale} $$
For quanto derivatives we prefer to express the equity/forex dynamics in terms of $X_t$ the FOR/DOM exchange rate instead of the DOM/FOR exchange rate $\tilde{X}_t$. This can be done through a simple application of Itô's lemma noticing that $\tilde{X}_t = 1/X_t$. This would typically yield:
\begin{align}
\frac{dX_t}{X_t} = (r_d - r_f) dt + \sigma_X dW_t^{X,\mathbb{Q}^d} \\
\frac{dY_t}{Y_t} = (r_f - \rho_{XY}\sigma_X\sigma_Y) dt + \sigma_Y dW_t^{Y,\mathbb{Q}^d}
\end{align}
where we have introduced
$$ W_t^{X,\mathbb{Q}^d} = -W_t^{\tilde{X},\mathbb{Q}^d} $$
such that
$$ \langle W_t^{X,\mathbb{Q}^d}, W_t^{Y,\mathbb{Q}^d} \rangle_t = \rho_{XY} t = -\rho t $$
and we used $\sigma_X = \sigma_{\tilde{X}}$ for clarity.
Hence finally, under $\mathbb{Q}^d$ we can write:
$$ X_t = X_0 \exp \left(\left(r_d-r_f-\frac{\sigma_X^2}{2}\right)+\sigma_X W_t^{X,\mathbb{Q}^d} \right) $$
$$ Y_t = Y_0\exp \left(\left(r_f - \rho_{XY}\sigma_X\sigma_Y -\frac{\sigma_Y^2}{2}\right)t + \sigma_Y W_t^{Y,\mathbb{Q}^d} \right) $$
where the quantity
$$ F(0,t) = E^{\mathbb{Q}^d}_0 \left[ Y_t \right] = Y_0\exp \left(\left(r_f - \rho_{XY}\sigma_X\sigma_Y\right)t\right) $$
is known as the quanto forward.
and it is once again easy to show that
$$ \frac{Y_tX_t}{B_t^d} \text{ is a } \mathbb{Q}^d \text{- martingale} $$
using the fact that $Z=Y_t X_t$ is a product of lognormals (and not a ratio as before)
