Let $(\Omega, (\mathcal F, \mathbb P)$ be a probability space, $T$ some index set, and $(S, \Sigma)$ a measurable space. Let $X^f$ and $X^g : T × \Omega \to S$ be stochastic processes defined by the stochastic differential equations
Let $$dX^{f}_t = f(X_t) dt + \sigma(X_t)dW_t, \\ dX^{g}_t = g(X_t) dt + \sigma(X_t)dW_t.$$
Denote by $\mathbb Q_f$, resp. $\mathbb Q_g$, the corresponding laws, i.e.,
$$\mathbb Q_f[A] = \mathbb P[\{\omega: t\mapsto X_t^f(\omega) \in A \}] $$
I want to compute the Radon-Nikodym derivative $\frac {d\mathbb Q_f}{ d \mathbb Q_g}$.
I know that the answer is connected to Girsanov's theorem, which needs to be applied twice.
Girsanov's Theorem Let $\theta_t$ be an adapted process satiysfying the Novikov condition. Define a probability measure $\mathbb Q$ by $\frac{d\mathbb Q}{d\mathbb P}\Big|_{\mathcal F_t} = \mathcal E(\theta_t)$. Then the stochastic process $t \mapsto W_t - \int_0^t \theta_s ds$ is a standard Wiener process under the measure $\mathbb Q$.
My (updated) attempt.
As pointed out by @zhoraster, I should consider the process $Y_t = \Sigma^{-1}(X^f_t) X^f_t$, which satisfies
$$dY^f_t = \sigma^{-1}(X_t) f(X_t) dt + dW_t\enspace,$$
i.e. $Y:t\mapsto \int_0^t \sigma^{-1}(X_s) f(X_s) ds + W_t$.
According to Girsanov's theorem, this is a standard Wiener process under $\mathbb Q$, defined by $\frac{d\mathbb Q}{d\mathbb P} = \mathcal E\left(-\sigma^{-1}(X_t) f(X_t)\right)$, i.e., with $\theta_t = -\sigma^{-1}(X_t) f(X_t)$.
Further, $Y_t^0$ is a standard Wiener process wrt to $\mathbb Q$.
Define $$\Phi: \Omega \to C([0,T],S)\\ \omega \mapsto \left(t\mapsto W_t(\omega)\right) \enspace.$$
We know (?) that the laws $\mathbb Q_{Y_t^f}$ and $\mathbb Q_{Y_t^0}$ satisfies for all measurable $A \subset C(T,S)$ that $$ \mathbb Q_{Y_t^0}[A] = \mathbb P[\{\omega: t\mapsto W_t(\omega) \in A \}] = \mathbb P[\Phi^{-1}(A)] = (\Phi_* \mathbb P) [A] \\ \mathbb Q_{Y_t^f}[A] = \mathbb Q[\{\omega: t\mapsto W_t(\omega) \in A \}] = \mathbb Q[\Phi^{-1}(A)] = (\Phi_* \mathbb Q) [A] $$
By the definition of $\mathbb Q$, this means that $$\mathbb Q_{Y_t^f}\big[A\big] = \mathbb Q\left[\Phi^{-1}(A)\right] = \mathbb E_{\mathbb P}\left[\mathcal E(\theta_t) \mathbb 1_{\Phi^{-1}(A)}\right] $$
In other words, $\mathbb Q_{Y_t^0} = \Phi_* \mathbb P$ and $\mathbb Q_{Y_t^f} = \Phi_* \mathbb Q$. If the former is true, then my question reduces to whether the Radon-Nikodym derivative of push forward measures is the same as the Radon-Nikodym derivative of the original measures. Presumably, this only holds if $\Phi$ satisfies certain properties.
Update: Bounty added