First, before addressing your questions, let's prove slightly different form of CRT, one that is very easy to remember and often more efficient to apply.
Theorem $\:$ (Easy CRT) $\rm\ \ $ If $\rm\ m,\:n\:$ are coprime integers then $\rm\ m^{-1}\ $ exists $\rm\ (mod\ n)\ \ $ and
$\rm\displaystyle\quad\quad\quad\quad\quad \begin{eqnarray}\rm x&\equiv&\rm\ a\ (mod\ m) \\
\rm x&\equiv&\rm\ b\ (mod\ n)\end{eqnarray} \ \iff\ \ x\ \equiv\ a + m\ \bigg[\frac{b-a}{m}\ mod\ n\:\bigg]\ \ (mod\ m\:n)$
Proof $\rm\ (\Leftarrow)\ \ \ mod\ m:\ x\ \equiv\ a + m\ [\,\cdots\,]\ \equiv\ a\:,\ $ and $\rm\ mod\ n\!\!:\ x\ \equiv\ a + (b-a)\ m/m\ \equiv\ b\:.$
$\rm (\Rightarrow)\ \ $ The solution is unique $\rm\ (mod\ m\:n)\ $ since if $\rm\ x',\:x\ $ are solutions then $\rm\ x'\equiv x\ $ mod $\rm\:m,n\:$ therefore $\rm\ m,\:n\ |\ x'-x\ \Rightarrow\ m\:n\ |\ x'-x\ \ $ since $\rm\ \:m,\:n\:$ coprime $\rm\:\Rightarrow\ lcm(m,n) = m\:n\:.\ \, $ QED
Note $\ $ Easy CRT is not only easy to apply, but also very easy to remember. Namely note $\rm\ x\equiv a\pmod m\iff x = a + m\,k,\:$ for some integer $\rm\:k\:$. This further satisfies the second congruence iff $\rm\:mod\ n\!:\ x = a + m\,k\equiv b$ $\iff$ $\rm
k\:\equiv (b-a)/m,\ $ hence the "Easy CRT" solution. This enables the $(\Leftarrow)$ proof, i.e. fill in the dots in $\rm\:a + m\ [\,\cdots\,]\:$ so that it is $\rm\equiv b\pmod n\:$
The extended Euclidean algorithm may be used to compute the modular inverse $\rm\, m^{-1}\pmod n $ and, hence, the above bracketed value $\rm\ (b-a)/m = (b-a)m^{-1}\ mod\ n.\ $ For small numbers one can often compute this "fraction" by judicious cancellation and twiddling, e.g. see the many worked examples in prior posts here.
The more symmetric form of CRT that you mention in your question often proves more convenient for theoretical (vs. computational) purposes. Notice that it works by computing the elements $\rm\,e = uq \equiv 1\pmod p\,$ and $\rm\,f = vp\equiv 1\pmod q,\, $ i.e. $\rm\, e \equiv (1,0),\, f \equiv (0,1)\ mod\ (p,q).\,$ Then $\rm\,y = (a,b) = a(1,0)+b(0,1) = ae+bf\,$ yields the solution. This can be presented more structurally as a ring isomorphism $\rm\,\Bbb Z/pq \cong \Bbb Z/p \times \Bbb Z/q,\,$ as rschwieb touches on in his answer.
Note that $\rm\,e,f\,$ are idempotents, i.e. $\rm\,e^2\!\equiv e,\,f^2\!\equiv f\,$ by $\rm \,e^2\! = (1,0)^2\! = (1^2,0^2)= (1,0)$. Generally idempotents in $\rm\:\Bbb Z_n\:$ correspond to factorizations of $\rm\:n\:$ into coprime factors. Namely, if $\rm\:e^2 = e\in\Bbb Z_n\:$ then $\rm\:n\:|\:e(e\!-\!1)\:$ so $\rm\:n = jk,\ j\:|\:e,\ k\:|\:e\!-\!1,\:$ so $\rm\:(j,k)= 1\:$ by $\rm\:(e,e\!-\!1) = 1.\:$ Conversely if $\rm\:n = jk\:$ for $\rm\:(j,k)= 1,\:$ then by CRT, $\rm\:\Bbb Z_n\cong \Bbb Z_j\times \Bbb Z_k\:$ which has nontrivial idempotents $\rm\:(0,1),\,(1,0).\:$ It is not difficult to explicitly work out the details of the correspondence. Some integer factorization algorithms can be viewed as searching for nontrivial idempotents. For more on the correspondence between idempotents and factorizations see Peirce Decomposition.
Best Answer
The Chinese Remainder Theorem approach actually leads you to do more work than directly applying Euclid's algorithm in this simple case, because you now need to solve 3 congruence and apply Euclid/Bezout twice (on the moduli $2,11,17$, although you can avoid doing the 2 by inspection instead of via Bezout) to match them up.
If you can just apply Euclid directly: \begin{align*} 374-19\times 19&=13\\ 19-13&=6\\ 13-2\times 6&=1 \end{align*} and so running it backwards gives $3\times 374-59\times 19 = 1$. Hence multiplying your given equation by $-59$ gives $x\equiv -59\times 7\pmod{374}$.
On the other hand, by Chinese Remainder Theorem, you need to solve $$ \left\{ \begin{aligned} 19 x&\equiv 7\pmod{2}\\ 19 x&\equiv 7\pmod{11}\\ 19 x&\equiv 7\pmod{17}\\ \end{aligned} \right. $$ giving (steps omitted here) $$ \begin{aligned} x&\equiv 1\pmod 2\\ x&\equiv 5\pmod{11}\\ x&\equiv 12\pmod{17} \end{aligned} $$ and now you need to apply Euclid to $11,17$ and run backwards, giving $$ 2\times 17-3\times 11=1 $$ so $x\equiv 5\times (2\times 17)+12\times(-3\times 11)\pmod{187}$ and $x\equiv 1\pmod 2$, so $$ x\equiv 5\times 2\times 17+12\times(-3)\times 11+187\pmod{374}. $$