How to read category diagrams? I have problems with very basic categorial reasoning.
Suppose we have a commutative "cone" diagram: $f:A \to B$, $g:B \to C$, $h:A \to C$
Is its "commutativity" equivalent of saying: $\forall x\in A \: (g(f(x)) = h(x))$?
If so, 
am I right that a diagram of a couple of parallel functions: $k:A \to B$, $m:A \to B$
does not commute if $k \neq m$
 A: The answer to you first question is: Yes, unless something different is said, the commutativity of such a diagram means $g\circ f=h$.
For the second part, the answer is usually no, I would say.
I think that there is no universal convention on what commutativity should mean for all possible diagrams. Sometimes one has to guess from the context. If you want to know for a specific article or book you are reading, you should include it in your question.
A: A usual definition of a diagram (careful, it is not Martin Bandenburg's one) is the following:

Definition. Let $\mathcal J$ be a small category and $\mathcal C$ be a category. A commutative diagram of shape $\mathcal J$ in $\mathcal C$ is a functor $\mathcal J \to \mathcal C$.

So, writting a diagram in the naive sense (a bunch of objects and arrows between them) in $\mathcal C$ and declare it commutative is not enough: one needs to specify the index category $\mathcal J$.
Let me explain why. A "naive diagram" in $\mathcal C$ like
$$ A \overset f {\underset g\rightrightarrows} B$$
could be a commutative diagram $D_0 \colon \mathcal J_0 \to \mathcal C$ or $D_1\colon \mathcal J_1 \to \mathcal C$ where 


*

*$\mathcal J_0$ is the category with objects $0,1$ and two distinct morphisms $i_0,i_1 \colon 0 \to 1$ (plus the identities); composition is the only possible one ; $\mathcal D_0$ then acts by
$$ 0 \mapsto A,\, 1\mapsto B,\, i_0 \mapsto f,\, i_1\mapsto g$$

*$\mathcal J_1$ is the category with objects $0,1,2$ and three  morphisms $i\colon 0\to 1,j \colon 1 \to 2, k\colon 0 \to 2$ (plus the identities) ; composition is the only one in which $k=ji$ ; and let $\mathcal D_1$ act by
$$ 0 \mapsto A,\, 1\mapsto A,\, 2 \mapsto B,\, i \mapsto \mathrm{id}_A,\, j\mapsto f, k\mapsto g$$


In the first case, the commutativity of the diagram does not implies $f=g$ (it is possible though). In the second case, the commutativity ensures $f=g$.
When the situation is ambiguous, just refer to the index category!
A: See here for a general, precise definition of a commutative diagram. It follows readily that a triangle
$$\begin{array}{cc} & B & \\ ~~~f\nearrow && \searrow g~~~~\\A & \underset{\large h}{\longrightarrow} & \overset{\phantom{-}}{B} \end{array}$$
commutes iff $h = g \circ f$, and that a diagram of two parallel morphisms
$$A~ \overset{\overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}} ~B$$
commutes iff $f=g$.
