However, each reaction at a different temperature has a different equilibrium constant. With that said, the amount of each molecule will change which also changes the concentration so the amount of NH3 would not necessarily be 2.
This is true. I have not watched the video, but if he is suggesting that the stoichiometric coefficients define the ratios of the equilibrium concentrations of the compounds, then he is wrong. Stoichiometric coefficients do play an important role in equilibrium expressions, but not that one.
He raised each concentration to the power of the starting coefficient for that molecule. If you start out with $\ce{3H2}$ and you need a wildly different concentration to reach equilibrium, why are we raising the concentration to the power of the starting amount of $\ce{H2}$?
This is how the equilibrium constant (and more broadly the reaction quotient) is defined.
Take your generic reaction:
$$\ce{aA + bB <=> cC + dD}$$
The reaction quotient $Q$ is defined as the product of the concentrations of the products divided by the product of the concentrations of the reactants. Since $\ce{a}$ equivalents of $\ce{A}$ are in the reactants, $[\ce{A}]$ is multiplied $\ce{a}$ times for $[\ce{A}]^{\ce{a}}$.
$$Q_c = \frac{\prod\lbrace[\text{Products}]_i\rbrace}{\prod\lbrace[\text{Reactants}]_i\rbrace} = \frac{[\ce{C}]^{\ce{c}}[\ce{D}]^{\ce{d}}}{[\ce{A}]^{\ce{a}}[\ce{B}]^{\ce{b}}} $$
The pressure version of $Q$ replaces concentrations with partial pressures $P_i$.
$$Q_p = \frac{\prod\lbrace (P_{\text{products}})_i\rbrace}{\prod\lbrace( P_{\text{reactants}})_i\rbrace} = \frac{P_{\ce{C}}^{\ce{c}} P_{\ce{D}}^{\ce{d}}}{P_{\ce{A}}^{\ce{a}} P_{\ce{B}}^{\ce{b}}}$$
The equilibrium constant ($K_c$ for concentrations and $K_p$ for pressures) is defined as the reaction quotient at equilibrium concentrations/pressures. At equilibrium:
$$Q_c = K_c = \frac{[\ce{C}]^{\ce{c}}[\ce{D}]^{\ce{d}}}{[\ce{A}]^{\ce{a}}[\ce{B}]^{\ce{b}}}$$
$$Q_p = K_p =\frac{P_{\ce{C}}^{\ce{c}} P_{\ce{D}}^{\ce{d}}}{P_{\ce{A}}^{\ce{a}} P_{\ce{B}}^{\ce{b}}}$$
The values of $K_c$ and $K_p$ are determined by experimentally measuring the concentrations or pressures of $\ce{A}$, $\ce{B}$, $\ce{C}$, and $\ce{D}$ and plugging them into the expressions above.
For the specific reaction of nitrogen and hydrogen to form ammonia, the equilibrium constant (at a given temperature) is determined by measuring the equilibrium concentrations/pressures and plugging them into the expressions below.
$$\ce{3H2(g) + 2N2(g) <=> 2NH3 (g)}$$
$$K_c = \frac{[\ce{NH3}]^3}{[\ce{H2}]^3 [\ce{N2}]^2}$$
$$K_p = \frac{P_{\ce{NH3}}^2}{P_{\ce{H2}}^3 P_{\ce{N2}}^2}$$
Once we know the value of $K_c$ or $K_p$, we can then use it and some partial concentration/pressure data to predict the equilibrium concentrations/pressures.
At 300 K, the $K_c$ of this process (the Haber Reaction) is $4.34\times10^{-3} \text{ M}^{-3}$. If we started with $[\ce{H2}]=3.0 \text{ M}$, $[\ce{N2}]=2.0 \text{ M}$, and $[\ce{NH3}]=2.0 \text{ M}$, we can calculate the equilibrium concentrations of all three species when the dust has settled. First we calculate $Q_c$ and compare to $K_c$. If $Q_c$ is greater than $K_c$, there is too much product, and the reaction will shift toward reactants. If $Q_c$ is less than $K_c$, there is too much reactant, and the reaction will shift toward products.
$$Q_c = \frac{[\ce{NH3}]^3}{[\ce{H2}]^3 [\ce{N2}]^2} = \frac{(2.0 \text{ M})^2}{(3.0 \text{ M})^3 (2.0 \text{ M})^2}= \frac{1}{27 \text{ M}^3}=3.70\times10^{-2} \text{ M}^{-3}$$
$$ 3.70\times10^{-2} > 4.34\times10^{-3} \space \therefore \space Q_c > K_c $$
At this combination, the reaction shifts toward reactants. $[\ce{H2}]$ and $[\ce{N2}]$ increase, and $[\ce{NH3}]$ decreases, all by a factor of $x$ multiplied by the stoichiometric coefficients of each species. The new concentrations are $[\ce{H2}]=3.0+3x \text{ M}$, $[\ce{N2}]=2.0+2x \text{ M}$, and $[\ce{NH3}]=2.0-2x \text{ M}$. The equilibrium expression is:
$$K_c 4.34\times 10^{-3} = \frac{(2-2x)^2}{(3+3x)^3 (2+2x)^2}$$
Since I chose the starting concentrations to be those values implied by the stoichiometry, I ended up (coincidentally) with a reaction near equilibrium. Thus, I cannot make assumptions about the value of $x$ relative to the concentrations, so no simplification is possible. To avoid mistakes, I will let WolframAlpha handle the algebra and solve for $x=3.18\times 10^{-1}$. The equilibrium concentrations are thus:
$$[\ce{H2}]_{eq} = 3+3(3.18\times 10^{-1})=3.95 \text{ M}$$
$$[\ce{N2}]_{eq}= 2+2(3.18\times 10^{-1})=2.64 \text{ M}$$
$$[\ce{NH3}]_{eq} = 2-2(3.18\times 10^{-1})=1.36 \text{ M}$$
Note that if we had started at different initial concentrations, we would end with different equilibrium concentrations. The maths are easier if you start with zero of one or two species.
I could also go on about the other way to determine the equilibrium constant (i.e. from the relationship $\Delta G^o=-RT\ln{K}$), but I have gone on long enough.
