The usage of commutative diagrams I'm starting to understand what commutative diagrams are, but I'm not sure about their purpose, what is their intended use and what kind of problems are solvable with them. By "solvable with a commutative diagram" I mean some fancy graphical reasoning, redrawing etc.
For example given only the commutative diagram for the exterior derivative
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
   \Omega^k(N)     & \ra{f^*}  &    \Omega^k(M)      \\
    \da{d}         &           &     \da{d}          \\
   \Omega^{k+1}(N) & \ras{f^*} &    \Omega^{k+1}(M)  \\
\end{array}
$$
is it even possible to tell that $d$ is a derivative, that is it is linear and the appropriate Leibniz rule holds?
Another example is my own, I may have done it totally wrong.
Given two vector spaces $V$ and $W$ (possibly of the same dimension) with scalar products, $f$ being a morphism, is it possible to prove that the following diagram commutes:
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
   V    & \ra{f}  &    W     \\
 \da{h} &         &  \da{h}  \\
   V    & \ras{f} &    W     \\
\end{array}
$$
only for $h$ of a certain form, which I guess is
$$h(v) = \lambda(v^2) v$$
where $\lambda$ is some arbitrary function? Is it a valid commutative diagram?
 A: A commutative diagram is a statement. In the case of your last diagram its claims that for all $v \in V$
$$
f(h_V (v)) = h_W (f(v)) \tag{1}
$$
where $h_{V} \colon V \to V$ is defined as 
$$h_V (v) = \lambda ( \langle v,v \rangle_V ) v
$$
where $\lambda$ is an arbitrary function $\lambda \colon \Bbb R \to \Bbb R$.
Substituting the definition of $h$ into (1) we have
$$
f\Big(\lambda( \langle v,v \rangle_V) v\Big) =  \lambda (\langle f(v),f(v) \rangle_W) f(v) \tag{2}
$$
Assuming that f is an isometry, that is f is linear and satisfies
$$\langle f(v),f(v) \rangle_W = \langle v,v \rangle_V
$$
so (2) becomes
$$
\lambda \langle v,v \rangle_V f(v) =  \lambda \langle v,v \rangle_V  f(v)
$$
Since we require $v$ be an arbitrary element of $V$ we may conclude that
$$
f(v) = f(v)
$$
for all $v \in V$.
Thus, the statement of your second diagram is true.
Edit. I corrected the above calculation to show that $\lambda$ can be an arbitrary real function, not just a multiplication to a scalar, as I erroneously assumed initially.
A: Commutative diagrams aren't a problem solving technique in the usual sense of the term. Rather, they're a language for efficiently describing certain kinds of relationships. You could remove the commutative diagrams from any proof that used them, but it would make the proof longer and harder to understand. (You could remove the commutative diagrams from the statements of any theorems that used them, too, but those would also become longer and harder to understand.) 
If you haven't run into a problem where it seems like using a commutative diagram would help you express some idea, then I wouldn't try to force the issue. Wait until you actually feel the need. 
In this particular case, the first commutative diagram you wrote down expresses the naturality of the exterior derivative. This is a useful thing to know: it guarantees that any computations you do involving exterior derivatives remain valid under pullback. But if you haven't run across a situation where it would be useful to know this, then wait. 
