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
Since the augmented filtration is right-continuous, we may assume that $(M_t)_{t \geq 0}$ has càdlàg sample paths. Since $(M_t)_{t \geq 0}$ is a local martingale, there is a sequence of stopping times $(\tau_k)$ such that $\tau_k \uparrow \infty$ and $(M_{t \wedge \tau_k})_{t \geq 0}$ is a martingale. Set
$$Y := M_{T \wedge \tau_k}$$
for fixed $k \in \mathbb{N}$ and $T>0$. Since $Y \in L^1$ and $L^2$ is dense in $L^1$, we can find a sequence $(Y_n)_{n \in \mathbb{N}} \subseteq L^2(\mathcal{F}_T)$ such that $Y_n \to Y$ in $L^1$. Obviously,
$$M_t^n := \mathbb{E}(Y_n \mid \mathcal{F}_t), \qquad t \leq T,$$
is an $L^2$-bounded $\mathcal{F}_t$-martingale. By the martingale representation theorem for $L^2$-martingales, there exists $\phi_n \in L^2(\lambda_T \otimes \mathbb{P})$ such that
$$M_t^n = \int_0^t \phi_n(s) \, dW_s, \qquad t \leq T.$$
In particular, $(M_t^n)_{t \leq T}$ has continuous sample paths. By the maximal inequality,
$$\mathbb{P} \left( \sup_{t \leq T} |M_{t \wedge \tau_k}-M_t^n| > \epsilon \right) \leq \epsilon^{-1} \mathbb{E}|Y-Y_n| \to 0,$$
i.e. $\sup_{t \leq T} |M_{t \wedge \tau_k}-M_t^n|$ converges in probability to $0$. Extracting a convergent subsequence, we conclude that $(M_{t \wedge \tau_k})_{t \leq T}$ has continuous sample paths. Since both $k$ and $T$ are arbitrary, we find that $(M_t)_{t \geq 0}$ has a.s. continuous sample paths. Now the claim follows using the argumentation described in the question.
For martingales with not necessarily continuous sample paths (which are not adapted to a filtration generated by a Brownian motion), we need more general representation results; the following result is due to Ikeda-Watanabe.
See also this question.