= Solution
The rational approximation below is taken in lowest terms, $(a,q)=1$, with $1\le q\le Q$, and $X>0$. This is necessary for denominator-sensitive <exponential sum> estimates and for the final <Ramanujan sum> to equal $\mu(q)$. If $0<X<1$, the defining sum over $a\le X$ is empty and $S_4=0$; hence we may assume $1\le X<\sqrt N$.
Set $c_u=\sum_{a\mid u,\ a\le X}\mu(a)$, so $|c_u|\le\tau(u)$, and write the given <Vaughan identity> term as the <Type II sum>
$$
S_4=-\sum_{\substack{u>X,\ d>X\\ud\le N}}c_u\Lambda(d)e(\theta ud),\qquad e(t)=e^{2\pi it}.
$$
Use a <dyadic decomposition> into $u\in(U,2U]$, $d\in(D,2D]$, retaining $ud\le N$. Empty blocks may be discarded; every remaining block has $U,D\ge X$, $UD<N$, and there are $O(\log^2(2N))$ blocks. Denote a block by $T$. The <Cauchy-Schwarz inequality> gives
$$
|T|^2\le\left(\sum_{U<u\le2U}|c_u|^2\right)
\sum_{U<u\le2U}\left|\sum_{\substack{D<d\le2D\\ud\le N}}\Lambda(d)e(\theta ud)\right|^2.
$$
By the allowed <divisor-square summatory bound>, the first factor is $O(U\log^3(2N))$. Expand the second factor. The diagonal $d_1=d_2$ contributes $O(UD\log^2(2N))$, using $0\le\Lambda(d)\le\log(2N)$. For a pair $d_1\ne d_2$, the allowed $u$ form the interval
$$
U<u\le\min(2U,N/d_1,N/d_2).
$$
Thus the <exponential geometric sum bound> applies to this interval even though the original block has a hyperbolic cutoff. For each difference $h=|d_1-d_2|\le D$, there are $O(D)$ ordered pairs, giving an upper bound
$$
\ll D\log^2(2N)\left(U+\sum_{1\le h\le D}\min(CU,\|\theta h\|^{-1})\right)
$$
for the expanded second factor, with an absolute constant $C$ accommodating integer endpoints. The supplied <reciprocal fractional-part sum near a rational>, with $M\asymp D$ and $R\asymp U$, bounds the last sum by
$$
O\left((UD/q+U+q+D)\log(2N)\right).
$$
One may also derive this directly from the <separated reciprocal-distance sum> by splitting $h$ into blocks of length at most $q/2$; the approximation error between two block points is at most $1/(2q)$, and the rational rotations are separated by $1/q$.
Combining the estimates yields
$$
|T|^2\ll UD\left(UD/q+U+q+D\right)\log^6(2N).
$$
Since $UD<N$ and $U,D\ge X$, both $U,D\le N/X$. Taking square roots proves
$$
|T|\ll\left(\frac N{\sqrt q}+\frac N{\sqrt X}+\sqrt{Nq}\right)\log^3(2N).
$$
Summing the dyadic blocks establishes the <rational-phase Type II estimate> and, in particular, the requested bound with an explicit logarithmic power:
$$
\boxed{|S_4|\ll\left(\frac N{\sqrt q}+\frac N{\sqrt X}+\sqrt{Nq}\right)\log^5(2N).}
$$
The <Siegel–Walfisz theorem> states that for fixed $A,C>0$, uniformly for $q\le(\log N)^A$ and $(b,q)=1$,
$$
\psi(N;q,b):=\sum_{\substack{n\le N\\n\equiv b\pmod q}}\Lambda(n)
=\frac N{\varphi(q)}+O_{A,C}\left(N(\log N)^{-C}\right).
$$
The constant is allowed to be ineffective. Apply this with $C=A+B+2$ to the reduced residue classes. Their contribution to the prime <exponential sum> is
$$
\sum_{\substack{n\le N\\(n,q)=1}}\Lambda(n)e(an/q)
=\frac N{\varphi(q)}\sum_{\substack{1\le b\le q\\(b,q)=1}}e(ab/q)
+O_{A,B}\left(N(\log N)^{-B}\right),
$$
since there are at most $q\le(\log N)^A$ residue classes. To evaluate the <Ramanujan sum>, the <Möbius divisor-sum identity> gives $\mathbf1_{(b,q)=1}=\sum_{d\mid(b,q)}\mu(d)$. Then the <finite geometric series> gives
$$
\begin{aligned}
c_q(a)&=\sum_{d\mid q}\mu(d)\sum_{v=1}^{q/d}e(av/(q/d))\\
&=\sum_{\substack{m\mid q\\m\mid a}}m\mu(q/m).
\end{aligned}
$$
When $(a,q)=1$, only $m=1$ remains, giving $c_q(a)=\mu(q)$. The omitted nonunits contribute only <prime powers> $p^j$ with $p\mid q$. For each such prime, the sum of their <Von Mangoldt function> weights is at most $\log N$, so their total is at most $\omega(q)\log N=O_A(\log N\log\log(3N))$, absorbed by the stated error. This proves the <major-arc value of the prime exponential sum>:
$$
\boxed{\widehat\Lambda_N(a/q)=\frac{\mu(q)}{\varphi(q)}N
+O_{A,B}\left(N(\log N)^{-B}\right)\quad((a,q)=1).}
$$
The original PDF does not state coprimality in this last request. Without it the asserted main term is genuinely false: $a=q=2$ makes $e(an/q)=1$ and hence $\widehat\Lambda_N(a/q)=\psi(N)\sim N$, whereas $\mu(2)/\varphi(2)=-1$. For an arbitrary numerator the correct main term is $N c_q(a)/\varphi(q)$, with $c_q(a)$ given by the divisor formula above. Reducing the fraction before applying the boxed result is the equivalent repair.
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