stochastic processes

Assume that $E \int_0^t f^2(W_u)du < \infty.$ Define \begin{align} I_t &:= \int_0^t f(W_u)dW_u, 0\leq t \leq T;\\ X_t &:= \int_0^t f^2(W_u)du, 0\leq t \leq T. \end{align} Then \begin{align} E[I_t^2-X_t | \mathscr F_s] &= E\left[\left( \int_0^t f(W_u)dW_u \right)^2-\left. \int_0^t f^2(W_u)du \right| \mathscr F_s \right] \\ &= E\left[I_s^2 + \left( \int_s^t f(W_u)dW_u \right)^2 +2I_s \int_s^t f(W_u)dW_u -X_s - \left. \int_s^tf^2(W_u)du \right | \mathscr F_s \right]\\ &= I_s^2 + E\left[ \left(\int_s^t f(W_u)dW_u \right)^2 \right]+2I_s \underbrace{E\left[ \int_s^t f(W_u)dW_u \right]}_{=0} -X_s \\ &- E\left[ \int_s^t f^2(W_u)du \right] \\ &=I_s^2 - X_s, \end{align} where in the last step I have used the Ito isometry. Thus the process $I^2 -X$ is a martingale and therefore a local martingale and by definition of quadratic vatiation (your definition in the question), X is the quadratic variation process of I.

The other question, the process $\int_0^t f(W_u)du, 0 \leq t \leq T$ is of finite variation so its quadratic variation is zero.