In the page 16 of the book "The Malliavin Calculus and Related Topics" from Nualart one reads:
Why do we need $u$ to be progressively measurable to ensure that $\tilde u$ is adapted?
Is there an example of non progressively measurable process $u$ such that $\tilde u^n$ as in $1.17$ is well defined (meaning, that it makes sense to integrate $u$ in the intervals $(\frac{i-1}{2^n},\frac{i}{2^n})$ ) and $\tilde u^n$ is not adapted?
The progressive measurability of $u$ ensures that $\int_{(i-1)2^{-n}}^{i2^{-n}}u(s)\,ds$ is $\mathcal F_{i2^{-n}}$ measurable, and this implies that $\tilde u_n(t)$ is $\mathcal F_t$ measurable (because the only terms in the sum for $\tilde u_n(t)$ that contribute a non-zero amount are those for which $i2^{-n}<t$). [N.B. Nualart's citation concerning progressively measurable modifications is wrong; it should read [224, Theorem IV.46].]
As to your other question, see the Answer in [https://mathoverflow.net/questions/176622/progressively-measurable-vs-adapted]. The reference there to Example 1.38 of M. Scheutzow [http://page.math.tu-berlin.de/~scheutzow/WT3main.pdf] provides an example that may be helpful to your understanding. Scheutzow's example gives a process $u$ (the indicator of his $A$) that is measurable, adapted, but not progressive. For this $u$, Nualart's approximation $\tilde u_n(t)$ is identically $0$. The progressive modification promised in Nualart's reference is the identically 0 process.
Suppose the filtered probability space $(\Omega,\mathcal F,(\mathcal F_t)_{t\ge 0},\Bbb P)$ we are working with is "complete" in the sense that $(\Omega,\mathcal F,\Bbb P)$ is a complete measure space and $\mathcal F_0$ contains all the $\mathcal F$ null sets. (This is part of the "usual conditions".) Let $u$ be an element of $L_a^2(T\times\Omega)$ and let $v\in L^2_a(T\times \Omega)$ be a progressively measurable modification of $u$. To keep it simple, let's even assume that $|u(t,\omega)|+|v(t,\omega)|\le K$ for all $(t,\omega)$, where $K$ is some constant. Because $v$ is a modification of $u$, Fubini's theorem tells us that for $\Bbb P$-a.e. $\omega\in\Omega$, the set $\{s\ge 0: v(s,\omega)\not=u(s,\omega)\}$ is Lebesgue null. It follows that for $\Bbb P$-a.e. $\omega$, you have $\tilde v_n(t)=\tilde u_n(t)$ for all $t\ge 0$. (Here $\tilde v_n(t)$ is obtained by applying Nualart's recipe to $v$ rather than $u$.) Because $v$ is progressive, $\tilde v_n(t)$ is $\mathcal F_t$ measurable for each $t$. Because the filtration is complete, it follows that $\tilde u_n(t)$ is also $\mathcal F_t$ measurable. Thus, assuming only that $u$ is measurable and adapted, you have that $\tilde u_n$ is adapted for each $n$, provided the filtration is complete. So in the complete case, the difference between "measurable and adapted" and "progressive" can be safely ignored.