Without your (detailed) calculations it is hard to say what you did wrong. So let's do it step by step:
It is not difficult to check that $(W_t^2-t)$ is a martingale. Therefore, we have
$$\begin{align*} \mathbb{E}(t W_t^2 \mid \mathcal{F}_s) &= t \mathbb{E}((W_t^2-t)+t) \mid \mathcal{F}_s) \\ &= t (W_s^2-s+t) = t W_s^2 + t^2-ts. \end{align*}$$
That agrees with your result for this term. Now the next one.
$$\begin{align*} \mathbb{E} \left( \int_0^t W_u^2 \, du \mid \mathcal{F}_s \right) &= \int_0^s W_u^2 \, du + \mathbb{E} \left( \int_s^t W_u^2 \, du \mid \mathcal{F}_s \right) \\ &= \int_0^s W_u^2 \, du + \mathbb{E} \left( \int_s^t ((W_u-W_s)+W_s)^2 \, du \mid \mathcal{F}_s \right) \\ &= \int_0^s W_u^2 \, du + \int_s^t \mathbb{E}((W_u-W_s)^2) \, du + 2W_s \int_s^t \underbrace{\mathbb{E}(W_u-W_s)}_{0} \, du \\ &\quad + W_s^2 (t-s) \\ &= \int_0^s W_u^2 \, du + \int_s^t (u-s) \, du + W_s^2 (t-s) \\ &= \int_0^s W_u^2 \, du + \frac{(t-s)^2}{2} + W_s^2 (t-s). \end{align*}$$
This is different from your result. The third one is again correct. Adding all up, we get
$$\begin{align*} \mathbb{E}(X_t \mid \mathcal{F}_s) &= t W_s^2 + t^2-ts - \left( \int_0^s W_u^2 \, du + \frac{(t-s)^2}{2} + W_s^2 (t-s) \right) - \frac{t^2}{2} \\ &= s W_s^2 - \int_0^s (W_u^2-u) \, du = X_s \end{align*}$$
For brevity, set
$$q_t := \exp \left( \int_0^t f(s) \,d W_s - \frac{1}{2} \int_0^t f(s)^2 \, ds \right),$$
i.e. $d\mathbb{Q} = q_t \, d\mathbb{P}$. Since
$$W_t = \tilde{W}_t + \int_0^t f(s),$$
we have
\begin{align*} q_t &= \exp \left( \int_0^t f(s) dW_s + \int_0^t f(s) (f(s) \, ds) - \frac{1}{2} \int_0^t f(s)^2 \,ds \right) \\ &= \exp \left( \int_0^t f(s) \, d\tilde{W}_s + \frac{1}{2} \int_0^t f(s)^2 \, ds \right). \tag{1} \end{align*}
From
$$\mathbb{P}(\tilde{W}_t \in A) = \int 1_{\{\tilde{W}_t \in A\}} \underbrace{d\mathbb{P}}_{=1/q_t \, d\mathbb{Q}} = \int 1_{\{\tilde{W}_t \in A\}} \frac{1}{q_t} \, d\mathbb{Q}$$
we see that
$$\mathbb{P}(\tilde{W}_t \in A) \stackrel{(1)}{=} \int 1_{\{\tilde{W}_t \in A\}} \exp \left( -\int_0^t f(s) \, d\tilde{W}_s - \frac{1}{2} \int_0^t f(s)^2 \, ds \right) \, d\mathbb{Q}.$$
By Girsanov' theorem, $(\tilde{W}_s)_{s \leq t}$ is a Brownian motion with respect to $\mathbb{Q}$, i.e. it has the same distribution as $(W_s)_{s \leq t}$ with respect to $\mathbb{P}$. Consequently,
$$\mathbb{P}(\tilde{W}_t \in A) = \int 1_{\{W_t \in A\}} \exp \left( - \int_0^t f(s) \, dW_s - \frac{1}{2} \int_0^t f(s)^2 \, ds \right) \, d\mathbb{P}.$$
Best Answer
Proof: Fix a deterministic function $f \in L^2(0,T)$. By the denseness of the continuous functions, there exists a sequence $(f_n)_{n \in \mathbb{N}}$ of continuous functions such that $f_n \to f$ in $L^2(0,T)$, i.e. $$\lim_{n \to \infty} \int_0^T |f_n(t)-f(t)|^2 \, dt = 0. \tag{1}$$ By Itô's isometry, this implies that the stochastic integral $X_n := \int_0^T f_n(s) \, dW_s$ converges in $L^2(\mathbb{P})$ to $X:=\int_0^T f(s) \, dW_s$. Since $f_n$ is continuous, we know that $X_n$ is Gaussian with mean $\mu_n =0$ and variance $\sigma_n^2 = \int_0^T f_n(s)^2 \, ds$. As $\lim_{n \to \infty} \mu_n=0$ and $\lim_{n \to \infty} \sigma_n^2 = \int_0^T f(s)^2 \, ds$, it follows that the limit $X=\lim_{n \to \infty} X_n$ is Gaussian with mean zero and variance $\int_0^T f(s)^2 \, ds$ (this can be shown e.g. by using that the characteristic function of $X_n$ converges pointwise to the characteristic function of $X$).
For continuous deterministic functions $f$, the stochastic integral can be obtained as ($L^2$-)limit of Riemann sums: $$\int_0^T f(s) \, dW_s = \lim_{n \to \infty} \sum_{i=0}^{n-1} f \left( T\frac{i}{n} \right) (W_{T(i+1)/n}-W_{Ti/n}). \tag{2}$$ Using that the increments $W_{T(i+1)/n}-W_{Ti/n}$ are independent and distributed like $N(0,T/n)$, we find that
\begin{align*} \mathbb{E}\exp \left( i \xi \int_0^T f(s) \, dW_s \right) &= \lim_{n \to \infty} \prod_{i=0}^{n-1} \mathbb{E}\exp \left( i \xi f(Ti/n) (W_{T(i+1)/n}-W_{Ti/n}) \right) \\ &= \lim_{n \to \infty} \prod_{i=0}^{n-1} \exp \left( - \xi^2 \frac{f(Ti/n)^2}{2} \frac{T}{n} \right) \\ &= \lim_{n \to \infty} \exp \left(-\xi^2 \sum_{i=0}^{n-1} f(Ti/n)^2 \left[ \frac{T(i+1)}{n} - \frac{Ti}{n} \right] \right) \\ &= \exp \left( - \frac{\xi^2}{2} \int_0^T f(s)^2 \,d s \right). \end{align*}
This shows that $\int_0^T f(s) \, dW_s$ is Gaussian with mean zero and variance $\int_0^T f(s)^2 \, ds$.