Deriving Chinese Remainder Theorem from gcd Bezout identity We want to find the solution $x$ to the congruence system
$$\begin{align} x &\equiv r_1 \!\!\!\pmod{\!m_1}\\
x &\equiv r_2 \!\!\!\pmod{\!m_2}\end{align},\ \ {\rm where}\ \  \gcd(m_1, m_2) = 1$$
These can be rewritten as
$$\begin{align} x &= r_1 + m_1 j\\ x &= r_2 - m_2k\end{align}$$
for unknown integers $j, k$. So we set those two equations as equal and rearrange them to
$$m_1j + m_2k = r_2 - r_1$$
Now assume we perform $\text{egcd}(m_1, m_2)$, which gives us the solution to
$$m_1u + m_2v = \gcd(m_1, m_2)$$
Let the integer $\,h\,$ make $\,h \gcd(m_1, m_2) = r_2 - r_1$.
Then $\qquad hm_1u + hm_2v = h\gcd(m_1, m_2) =  r_2 - r_1 = m_1j + m_2k$
At this point I am lost because I don't think we can just assume  $hm_1u = m_1j$ and extract $j = hu$. 
How do I get the values of $j$ or $k$ so I can get the value of $x$?
 A: Correct, if $x$ is a root of the congruences then $\, x= j\,m_1 + r_1 = -k\, m_2 + r_2\,$ has roots $\,j,k\in\Bbb Z$.
This argument reverses:$ $ if $\,j,k\,$ are integers with $\ j\, \color{#c00}{m_1}+ k\, \color{#0a0}{m_2} =\, \color{#0a0}{r_2} - \color{#c00}{r_1}\, $ then by $\rm\color{#c00}{re}\color{#0a0}{arranging}$ $\  x :=\ \color{#c00}{r_1} +\, j\ \color{#c00}{m_1}^{\phantom{|}}\ =\,\ \ \color{#0a0}{r_2}\: -\,\ k\ \color{#0a0}{m_2}\ $ is one solution  of the given system of congruences because $\,x\equiv \color{#c00}{r_1}\!\!\pmod{\!\!\color{#c00}{m_1}}^{\phantom{|^|}}\!\!\!,$ $x\equiv \color{#0a0}{r_2}\!\pmod{\!\!\color{#0a0}{m_2}}.\,$ Since you've already discovered one such solution for $\,j,k,\,$ viz.  $\,j=hu,\,k^{\phantom{|^|}}\!\! = hv,\,$  you need only substitute into the above rearranged CRT solution for $\,x.$
Remark $ $ Combining both directions above and adding a final gcd equivalence yields the following
CRT Solvability Criterion $ $ (which immediately generalizes to ideals)
Theorem $\ \ \left.\exists\, x\in\Bbb Z\!:  \begin{align}x\equiv r_1\!\!\!\pmod{\!m_1}\\ x\equiv r_2\!\!\!\pmod{\!m_2}\end{align}\right\} \begin{array}{l}\!\iff  \exists\,j,k\in\Bbb Z\!:\ j\,m_1\! + k\, m_2 =\, r_2\!-r_1 \\ \!\iff\, \gcd(m_1,\,m_2)\mid r_2 -r_1\end{array}$
Proof $ $ Clearly $\,d := \gcd(m_1,m_2)\mid r_2-r_1 \,$ is a necessary condition  for the equation to have roots $\,j,k\in \Bbb Z,\,$ by $\,d\mid m_1,m_2\Rightarrow\, d^{\phantom{|}}_{\phantom{i}}\!\mid j m_1\! + km_2 = r_2 - r_1.\,$ Further this condition is also sufficient by Bezout (or, constructively, by the extended Euclidean algorithm), i.e. we can scale the Bezout equation $\, a m_1\! + b m_2 = d\,$ by $\, c = \large \frac{r_2\,-\,r_1^{\phantom{.}}}{d}\,$ to get $\,ca\,m_1\!+cb\,m_2 = r_2-r_1 \,$ so, as above, rearranging this yields a congruence system solution: $\ x\, :=\, r_1 + ca\,m_1 = r_2 - cb\,m_2$.
Thus the congruence system is solvable $\iff d=\gcd(m_1,m_2)\mid r_2-r_1, \,$ i.e. iff the pair of congruences  is consistent mod their moduli gcd, and when true we can constructively read off a solution from the Bezout equation for the moduli by translating it into the equivalent system language as above, i.e. scale the Bezout equation to obtain the residue difference $\,r_1-r_2\,$ then rearrange it as above to obtain $\,x.\,$ Here is a worked example from this viewpoint, and here is some motivational examples. Summarizing the above method, we have derived the following very simple Bezout-based CRT (Chinese Remainder Theorem) method for solving congruence systems
$$\bbox[8px,border:2px solid #c00]{\text{$\color{#90f}{\text{scale}}$ the Bezout equation by the residue difference - then ${\rm \color{#c00}{re}\color{#0a0}{arrange}}$}}\qquad$$
$$\begin{align}
{\rm if}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\begin{array}{rr} &x\equiv\,  r_1\pmod{\!m_1}\\ &x\equiv\, r_2\pmod{\! m_2} \end{array}\ \ \ \ {\rm and}\ \ \ \ \gcd(m_1,m_2) = 1\\[.4em]
{\rm then}\ \ \  r_1 - r_2\, &=:\ \delta\qquad\qquad\qquad \rm residue\ difference \\[.2em] 
\times\qquad\quad\ \ \ 1\ &=\ \ a m_1\, +\, b m_2\quad\ \rm Bezout\ equation\ for \ \gcd(m_1,m_2) \\[.5em]\hline
\Longrightarrow\ \,r_1\ \color{#c00}{-\ r_2}\ \ \ \  &=\, \color{#0a0}{\delta am_1}\! + \delta bm_2\quad\ \rm product\ of \ above\ (= {\color{#90f}{scaled}}\ Bezout)\\[.2em]
\Longrightarrow\ \underbrace{r_1 \color{#0a0}{- \delta am_1}}_{\!\!\!\large \equiv\ r_1\! \pmod{\!m_1}}\! &=\,\ \ \underbrace{\color{#c00}{r_2}\ +\ \delta bm_2}_{\large\!\! \equiv\ r_2\! \pmod{\!m_2}}\ \ \ \  
 \underset{\large {\rm has\ sought\ residues}\phantom{1^{1^{1^{1^1}}}}}{\rm \color{#c00}{\!re}\color{#0a0}{arranged}\ product}\rm\! = {\small CRT}\ solution\end{align}\qquad$$
This generalizes: a system of  $n$ congruences is solvable $\iff$ each pair of congruences is solvable as above, and we can solve the system by successively replacing a pair of congruences by a  single equivalent congruence. By induction we eventually obtain a single congruence equivalent to the entire system.
Notice the above-linked ideal version can be expressed succinctly in terms of cosets, viz.
$$  \bbox[9px,border:2px solid #c00]{r_1\! +\! m_1\Bbb Z\,\cap\, r_2\! +\! m_2\Bbb Z \neq \phi \iff r_1-r_2 \in m_1\Bbb Z+m_2\Bbb Z}\qquad\qquad $$
