A filtration generated by Brownian motion simply means the smallest filtration with respect to which Brownian motion is adapted i.e. $$\mathcal{F}^B_t = \sigma \{B_s, \: s\leq t\}$$
On the other hand, a filtration for the Brownian motion, $\mathcal{F}_t$ is one to which the Brownian motion is adapted AND we have that for all $s<t$, $B_t - B_s$ is independent of $\mathcal{F}_s$. So notice that the filtration generated by the Brownian motion also satisfies these.
Now for your example. Let $X$ be a random variable independent of your Brownian motion. Then if we enlarge the Brownian filtration by adding in information about $X$: $$\mathcal{F}_t := \sigma \{X, B_s \: s\leq t\}$$ Then this is still a filtration for Brownian motion.
No, it's not exactly obvious. Let's prove the following theorem.
Let $(B_t)_{t \geq 0}$ be a one-dimensional Brownian motion and $(K_t)_{t \geq 0}$ a progressively measurable process such that $\mathbb{E}\int_0^t K_s^2 \, ds < \infty$ for all $t \geq 0$. If $(N_t)_{t \geq 0}$ is a continuous $L^2$-bounded martingale, then $$M_t := \int_0^t K_s \, dB_s$$ satisfies $$\langle M,N \rangle_t = \int_0^t K_s \, d\langle B,N \rangle_s.$$
Proof: Througout, $(N_t)_{t \geq 0}$ is an $L^2$-bounded martingale. First we consider the particular case that $K$ is a simple process of the form $$K(s) = \sum_{j=0}^{N-1} \varphi_j 1_{(s_j,s_{j+1}]}(s) \tag{1}$$ where $0<s_0 < \ldots < s_N$ and $\varphi_j \in L^2(\mathcal{F}_{s_j})$. We have to show that $M_t = \int_0^t K_r \, dB_r$ satisfies $$\mathbb{E}((M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s) = \mathbb{E} \left( \int_s^t K_r \, d\langle B,N \rangle_r \mid \mathcal{F}_s \right) \tag{2}$$ for any fixed $s \leq t$. Without loss of generality, we may assume that $s_N = t$ and that there exists $k \in \{0,\ldots,N\}$ such that $s_k = s$ (otherwise we refine the partition accordingly). Writing $$N_t-N_s = \sum_{i=k}^{N-1} (N_{s_{i+1}}-N_{s_i}) \quad \text{and} \quad M_t-M_s = \sum_{j=k}^{N-1} (M_{s_{j+1}}-M_{s_j})$$ we find
$$\mathbb{E}((M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s) = \sum_{j=k}^{N-1} \sum_{i=k}^{N-1} \mathbb{E}((M_{s_{j+1}}-M_{s_j}) (N_{s_{i+1}}-N_{s_i}) \mid \mathcal{F}_s).$$
Since both $(M_t)_{t \geq 0}$ and $(N_t)_{t \geq 0}$ are martingales, it is not difficult to see from the tower property that the terms on the right-hand side vanish for $i \neq j$, and so
$$\mathbb{E}((M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s) = \sum_{j=k}^{N-1} \mathbb{E}(\varphi_j (B_{s_{j+1}}-B_{s_j}) (N_{s_{j+1}}-N_{s_j}) \mid \mathcal{F}_s).$$
Using once more the tower property, we get
$$\begin{align*} \mathbb{E}((M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s) &= \sum_{j=k}^{N-1} \mathbb{E} \bigg[ \varphi_j \mathbb{E}((B_{s_{j+1}}-B_{s_j}) (N_{s_{j+1}}-N_{s_j}) \mid \mathcal{F}_{s_j}) \mid \mathcal{F}_s \bigg] \\ &= \sum_{j=k}^{N-1} \mathbb{E} \bigg[ \varphi_j \mathbb{E}(\langle B,N \rangle_{s_{j+1}}-\langle B,N \rangle_{s_j}) \mid \mathcal{F}_{s_j}) \mid \mathcal{F}_s \bigg]\\ &= \mathbb{E} \left( \sum_{j=0}^{N-1} \varphi_j (\langle B,N \rangle_{s_{j+1}}-\langle B,N \rangle_{s_j}) \mid \mathcal{F}_s \right) \\ &= \mathbb{E} \left( \int_s^t K_r \, d\langle B,N \rangle_r \mid \mathcal{F}_s \right) \end{align*}$$
This proves the assertion for the simple process $K$. For the general case we choose a sequence of simple process $(K_n)_{n \in \mathbb{N}}$ of the form $(1)$ such that $$\mathbb{E} \left( \int_0^t (K_n(s)-K(s))^2 \, ds \right) \to 0.$$ By the construction of the stochastic integral, this implies, in particular,
$$ \int_0^t K_n(r) \, dB_r \to \int_0^t K(r) \, dB_r \quad \text{in $L^2(\mathbb{P})$} \tag{3}$$
On the other hand, it follows from the Cauchy-Schwarz inequality that
$$\int_0^t K_n(r) \, d\langle B,N \rangle_r \to \int_0^t K(r) \, d\langle B,N \rangle_r \quad \text{in $L^2(\mathbb{P})$} \tag{4}$$
Thus, $M_t = \int_0^t K(r) \, dB_r$ satisfies
$$\begin{align*} \mathbb{E} \left( (M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s \right) &\stackrel{(3)}{=} \lim_{n \to \infty} \mathbb{E} \left[ \left( \int_0^t K_n(r) \, dB_r - \int_0^s K_n(r) \, dB_r \right) (N_t-N_s) \mid \mathcal{F}_s \right] \\ &\stackrel{(1)}{=} \lim_{n \to \infty} \mathbb{E} \left( \int_s^t K_n(r) \, d\langle B,N \rangle_r \mid \mathcal{F}_s \right) \\ &\stackrel{(4)}{=} \mathbb{E} \left( \int_s^t K(r) \, d\langle B,N \rangle_r \mid \mathcal{F}_s \right). \end{align*}$$
Best Answer
It might be easier to think about if we abstract a little. Brownian motion is just a continuous, time dependent random variable determined by some probability space $(X,\mathcal F, P)$. Let us view it as a function of two variables, $f(x,t)$. If $x$ is fixed, we have a continuous function, $f_x(t)=f(x,t)$, and we can compute it's integral $F(x,t)=F_x(t)=\int_0^t f_x(s) ds$, where we have just taken the usual integral of a continuous function (Riemann, Lebesgue, whatever, all the definitions agree for nice continuous functions). Thus, we have a transformation of the space of time dependent continuous random variables on $X$ (i.e., functions on $X\times \mathbb R$). The fact that we are working in particular with brownian motion doesn't enter into things.
A more complicated question is what it means to integrate a function or a random variable with respect to Brownian motion. This gets you to the Ito integral (and other similar variants) which are more subtle.