[proofplan]
We prove that averaging Dirichlet characters modulo $q$ gives exactly the indicator of the reduced residue class $a \pmod q$. Because $\gcd(a,q)=1$, the residue class $a$ contains only integers coprime to $q$, while every Dirichlet character vanishes on integers not coprime to $q$. Multiplying the resulting indicator identity by $\Lambda(n)$ and summing over $1 \leq n \leq x$ gives the desired decomposition.
[/proofplan]
[step:Prove the character average is the indicator of the residue class $a \pmod q$]
Let $G := (\mathbb{Z}/q\mathbb{Z})^\times$ be the reduced residue group modulo $q$, and let $\widehat{G}$ denote its group of complex-valued characters. For $n \in \mathbb{Z}$ with $\gcd(n,q)=1$, let $\bar{n} \in G$ denote the residue class of $n$, and let $\bar{a} \in G$ denote the residue class of $a$.
We claim that
\begin{align*}
\frac{1}{\varphi(q)}
\sum_{\chi \bmod q}
\overline{\chi(a)}\chi(n)
=
\begin{cases}
1, & n \equiv a \pmod q,\\
0, & n \not\equiv a \pmod q,
\end{cases}
\end{align*}
for every $n \in \mathbb{Z}$ with $\gcd(n,q)=1$.
Indeed, for such $n$, every Dirichlet character modulo $q$ restricts to an element of $\widehat{G}$, and
\begin{align*}
\overline{\chi(a)}\chi(n)
=
\chi(\bar{a})^{-1}\chi(\bar{n})
=
\chi(\bar{a}^{-1}\bar{n}).
\end{align*}
Thus
\begin{align*}
\frac{1}{\varphi(q)}
\sum_{\chi \bmod q}
\overline{\chi(a)}\chi(n)
=
\frac{1}{|G|}
\sum_{\chi \in \widehat{G}}
\chi(\bar{a}^{-1}\bar{n}).
\end{align*}
The finite character orthogonality relation on $G$ gives
\begin{align*}
\frac{1}{|G|}
\sum_{\chi \in \widehat{G}} \chi(g)
=
\begin{cases}
1, & g = e_G,\\
0, & g \neq e_G,
\end{cases}
\end{align*}
where $e_G$ is the identity element of $G$. Applying this with $g=\bar{a}^{-1}\bar{n}$ gives the displayed identity, since $\bar{a}^{-1}\bar{n}=e_G$ exactly when $n \equiv a \pmod q$.
[guided]
The goal is to turn a congruence condition into a character sum. Let
$G := (\mathbb{Z}/q\mathbb{Z})^\times$ be the group of invertible residue classes modulo $q$. Since $\gcd(a,q)=1$, the class $\bar{a}$ belongs to $G$. If $\gcd(n,q)=1$, then $\bar{n}$ also belongs to $G$.
For a Dirichlet character $\chi$ modulo $q$, the value $\chi(n)$ depends only on $\bar{n}$ when $\gcd(n,q)=1$. Also, because $\chi(a)$ is a root of unity, $\overline{\chi(a)}=\chi(a)^{-1}$. Hence
\begin{align*}
\overline{\chi(a)}\chi(n)
=
\chi(\bar{a})^{-1}\chi(\bar{n})
=
\chi(\bar{a}^{-1}\bar{n}).
\end{align*}
Therefore the average in question is an average of all characters on the single group element $\bar{a}^{-1}\bar{n}$:
\begin{align*}
\frac{1}{\varphi(q)}
\sum_{\chi \bmod q}
\overline{\chi(a)}\chi(n)
=
\frac{1}{|G|}
\sum_{\chi \in \widehat{G}}
\chi(\bar{a}^{-1}\bar{n}).
\end{align*}
The character orthogonality relation says that the average of all characters over a finite abelian group is $1$ at the identity element and $0$ away from the identity:
\begin{align*}
\frac{1}{|G|}
\sum_{\chi \in \widehat{G}} \chi(g)
=
\begin{cases}
1, & g = e_G,\\
0, & g \neq e_G.
\end{cases}
\end{align*}
Applying this with $g=\bar{a}^{-1}\bar{n}$ gives $1$ precisely when $\bar{n}=\bar{a}$, which is the same as $n \equiv a \pmod q$, and gives $0$ otherwise.
[/guided]
[/step]
[step:Extend the indicator identity to all positive integers]
For $n \in \mathbb{N}$ with $\gcd(n,q)>1$, every Dirichlet character modulo $q$ satisfies $\chi(n)=0$ by definition. Hence
\begin{align*}
\frac{1}{\varphi(q)}
\sum_{\chi \bmod q}
\overline{\chi(a)}\chi(n)
=
0.
\end{align*}
On the other hand, $n \not\equiv a \pmod q$, because $\gcd(a,q)=1$ while any integer congruent to $n$ modulo $q$ has a common divisor with $q$. Therefore, for every $n \in \mathbb{N}$,
\begin{align*}
\frac{1}{\varphi(q)}
\sum_{\chi \bmod q}
\overline{\chi(a)}\chi(n)
=
\begin{cases}
1, & n \equiv a \pmod q,\\
0, & n \not\equiv a \pmod q.
\end{cases}
\end{align*}
[/step]
[step:Multiply by $\Lambda(n)$ and sum over $1 \leq n \leq x$]
Multiplying the preceding identity by $\Lambda(n)$ and summing over all integers $n$ with $1 \leq n \leq x$, we obtain
\begin{align*}
\sum_{\substack{1 \leq n \leq x \\ n \equiv a \!\!\!\pmod q}} \Lambda(n)
&=
\sum_{1 \leq n \leq x}
\Lambda(n)
\left(
\frac{1}{\varphi(q)}
\sum_{\chi \bmod q}
\overline{\chi(a)}\chi(n)
\right)\\
&=
\frac{1}{\varphi(q)}
\sum_{\chi \bmod q}
\overline{\chi(a)}
\sum_{1 \leq n \leq x}
\chi(n)\Lambda(n).
\end{align*}
The interchange of sums is valid because both sums are finite. By the definitions of $\psi(x;q,a)$ and $\psi(x,\chi)$, this is exactly
\begin{align*}
\psi(x;q,a)
=
\frac{1}{\varphi(q)}
\sum_{\chi \bmod q}
\overline{\chi(a)}\,\psi(x,\chi).
\end{align*}
This proves the formula.
[/step]
