Since $f$ is deterministic, we get that $X_{t}$ is Gaussian https://math.stackexchange.com/questions/3221012/let-x-t-int-0t-sigma-tdb-t-what-is-the-law-of-x-t?noredirect=1&lq=1

with variance $\mathbb{E}\int_0^t f(s)^2 ds$.

This process is just a time changed Brownian motion see here https://almostsuremath.com/2010/04/20/time-changed-brownian-motion/

For the martingale part, this follows by the Ito construction increments $B_{t_{i}}-B_{t_{i-1}}$. That implies that the sum of them is also a martingale $I(f^{n}) = \int_{0}^{t} f^{n}_u dB_u$ where $f^{n}$ are simple processes.

Then finally we use that the $L_1$-limit of martingales is still martingale eg. see here When is the limit of Martingales a Martingale?

To get the formulas for the moments we use the Gaussianity. Since $f$ is deterministic, we get that $X_{t}$ is Gaussian with variance $\mathbb{E}\int_0^t f(s)^2 ds$. see here

The integral of a progressively measurable process $f$ is a limit of the integrals $I(f^{n}) = \int_{0}^{t} f^{n}_u dB_u$ where $f^{n}$ are simple processes and these integrals are Gaussian by definition.

Then as mentioned there we just use

Let $X_n$ be a sequence of normally distributed random variables with mean zero and variances $\sigma_n^2 \in [0,\infty)$. Suppose $X_n \to X$ in distribution. Then $\sigma_n^2$ converges to some $\sigma^2 \in [0,\infty)$, and $X$ is normally distributed with mean zero and variance $\sigma^2$.

In fact, this process is just a time changed Brownian motion see here