Derivative of a product and derivative of quotient of functions theorem: I don't understand its proof

The derivative product rule is a special case of the congruence product rule for rings, i.e.

Product Rule $\rm\quad\ a\equiv a_1,\; b\equiv b_1 \;\Rightarrow\;\; ab \equiv a_1 b_1$

Proof: $\rm\quad\; ab-a_1 b_1 \equiv a(b-b_1)+(a-a_1)b_1 \equiv 0 \quad$ QED

Thus $\rm\ \ f(x+t) \equiv \: f(x) + f'(x) \; t \;\pmod {t^2}$

and $\rm\ \ \ \,g(x+t) \equiv g(x) + g'(x) \; t \;\pmod {t^2}$

$\rm\ \ \Rightarrow\ \ f(x+t)g(x+t) \:\equiv\: f(x)g(x) + (f'(x)g(x) + f(x)g'(x)) \; t \:\;\pmod {t^2}$

$\rm\displaystyle\ \Rightarrow\ \ \frac{f(x+t)g(x+t)\: - \:f(x)g(x)}{t} \equiv\: f'(x)g(x) + f(x)g'(x) \quad\:\pmod t$

In fact this is how one universally defines derivatives in formal polynomial rings $\rm R[x]$, e.g. see here. This yields a purely algebraic approach to polynomial derivatives - devoid of limits or other topological notions.

The ring $\rm R[t]/t^2 \;$ is known as the algebra of dual numbers over the ring $\rm R$. This ring and its higher order analogs $\;\rm R[t]/t^n \;$ prove quite useful when studying (higher) derivations algebraically since they provide convenient algebraic models of tangent / jet spaces. E.g. as above, they permit easy transfer of properties of homomorphisms to derivations -- see for example section 8.15 in Jacobson, Basic Algebra II.

Dual numbers have been applied in many contexts, e.g. deformation theory [2], numerical analysis [3] (along with Levi-Civita fields), where they're viewed simply as truncated Taylor power series, and also in Synthetic Differential Geometry (SDG) [1], another rigorization of infinitesimals based on work of Lawvere and Kock. Note that SDG employs these nilpotent infinitesimals, unlike Abraham Robinson's nonstandard analysis, which employs invertible infinitesimals (hence contains infinite elements).

[1] Bell, J. L. Infinitesimals. Synthese 75 (1988) #3, 285--315.
http://www.jstor.org/stable/20116534

[2] Szendroi, B. The unbearable lightness of deformation theory, a tutorial introduction.
http://people.maths.ox.ac.uk/szendroi/defth.pdf

[3] M. Berz, Differential Algebraic Techniques,
in "Handbook of Accelerator Physics and Engineering, M. Tigner, A.Chao (Eds.)" (World Scientific, 1998)
http://bt.pa.msu.edu/cgi-bin/display.pl?name=dahape
http://bt.pa.msu.edu/NA/
http://bt.pa.msu.edu/pub/papers/

There is a more comprehensible (I hope) proof coming from thinking of derivative as a linear approximation.

By the definition of derivative¹, $f(x)=f(x_0)+f'(x_0)(x-x_0)+o(x-x_0)$ and $g(x)=g(x_0)+g'(x_0)(x-x_0)+o(x-x_0)$. Multiplying the equalities we get \begin{multline} f(x)g(x)=f(x_0)g(x_0)+(f'(x_0)g(x_0)+f(x_0)g'(x_0))(x-x_0)+\\ +f'(x_0)g'(x_0)(x-x_0)^2+o(x-x_0) \end{multline} and since $f'(x_0)g'(x_0)(x-x_0)^2=o(x-x_0)$, we can rewrite it as \[(fg)(x)=(fg)(x_0)+(f'g+fg')(x_0)\cdot(x-x_0)+o(x-x_o)\] — which exactly means that (fg)'=f'g+fg'.

¹ $\lim_{x\to x_0}g(x)=a$ iff $g(x)=a+o(1)$, hence $\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)$ iff $\frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)+o(1)$ which can be rewritten as $f(x)-f(x_0)=f'(x_0)(x-x_0)+o(x-x_0)$ (see e.g. wikipedia if "o(1)" notation is unfamiliar to you).

An equivalent way to state the product rule is $\frac{(fg)'}{fg} = \frac{f'}{f} + \frac{g'}{g}$. I prefer this statement because it is more intuitive: it says precisely that the relative instantaneous change in $fg$ is the sum of the relative instantaneous change in $f$ and the relative instantaneous change in $g$. In other words, this is just an expression of the approximation $(1 + a)(1 + b) \approx 1 + a + b$ when $a, b$ are both small ("multiplication near the identity is addition"). In other other words, this is an expression of the familiar fact that if you increase something by 5%, then by 5% again, the total increase is just a little more than 10%.

As for turning this into a formal proof, divide both sides of the expression you're confused about by $fg$. (This is not strictly valid if $f$ or $g$ is equal to zero, but I'm going for clarity here.)