You've entered into the wide and deep field of control theory. Here, Matlab with its comprehensive documentation will be your best guide. Matlab's help is a comprehensive compilation of many many textbooks. Where its help does not suffice, you can take a look at the references used to write the Help section you're interested in.
An impulse response is a unique way of describing a linear system. This means that two LTI systems with an identical impulse response can be judged to be mathematically identical - even if one is an automobile shock absorbeer and the other an electronic filter!
The impulse response is a useful way of testing the parameters of the system because it contains a wide range of frequencies. An ideal impulse contains ALL the frequencies from zero to infinity. In contrast, a sine wave has only a single frequency, so it's a very bad way of identifying a system. If your system is linear and time-invariant, a sinewave input will produce a sinewave on the output.
The problem you're describing is of a dual nature: you first identify a system. The knowledge of its parameters allows you to adjust the parameters of your controller. How do we work out these parameters? I recommend reading about System Identification in the Matlab Documentation. There are many textbooks on how to tackle this, but you're going to end up using a tool such as Matlab to learn this, anyway.
How do you adjust those control parameters? It's simpler than identifying a system, but there is a myriad of methods. There is no single way of doing this. There are dozens of controller topologies and many different ways of calculating the set of parameters that give you your desired response. For more information, read on about pole placement. There are many textbooks describing this. One of the best resources on this is the Matlab Control Toolbox Documentation.
For a simple SISO system you can derive the equations by hand. This tutorial explains more using a relatively simple example.
I hope this explains something.
Good question - it is a great example of a case where purely mathematical discussion can't shed a light on the issue.
You are correct about the first statement - causal impulse response of a system is an indication of causal system. The only correction is that it is if and only if statement, which also means that any causal system has causal impulse response.
The second statement is incorrect.
Mathematically speaking you're correct - the impulse response and the input are interchangeable inside the convolution integral, but there is more to this than just formalism.
I believe that your confusion arises from the simplified statement of causality: \$causality \Leftrightarrow \forall x[n]: y[n]=0, \forall n<0\$. This statement is correct, if you remember the underlying assumption, which is: \$x[n]=0, \forall n<0\$. In other words, this simplified statement is correct only for right sided inputs and appropriate choice of the origin of \$n\$.
The above can be stated in this way: if for each input that was zero before \$n=0\$ the output is also zero before \$n=0\$, then the system is causal.
The lack of symmetry between \$h[n]\$ and \$x[n]\$ in convolution integral arises from the fact that the impulse response is by definition a response to impulse at \$n=0\$, whereas the input is subject to a time shift - you can move the origin of \$n\$ which affects \$x[n]\$, but does not affect \$y[n]\$ (since the system is time invariant).
Best Answer
Assuming that with "associate" you mean "put in series" that is you hook S1 output to S2 input then you are not doing anything wrong. Problem is that sometimes (plenty of times) solving a convolution integral is not a piece of cake thus you prefere the Laplace transform.
Let's start by computing \$G_1(s)\$ and \$G_2(s)\$:
\$g_2(t)=e^{-\alpha t}u(t)\$, but \$U(s)=\mathcal{L}\{u(t)\}=\frac{1}{s}\$ and you can use the frequency shifting property that goes: $$ e^{at}f(t) \leftrightarrow F(s-a) $$ finally: $$ G_2(s)= \frac{1}{s+\alpha} $$
That's called exponential decay and is usually provided in Laplace transforms tables. Of course \$G_1(s)=G_2(s)|_{\alpha=1}\$ so that: $$ G_1(s) = \frac{1}{s+1} $$ Now we're searching for $$G(s)\triangleq G_1(s)G_2(s)=\frac{1}{s+1}\frac{1}{s+\alpha}=\frac{A}{s+1}+\frac{B}{s+\alpha}=\frac{s(A+B)+A\alpha+B}{(s+1)(s+\alpha)}$$ From the last step you can find out \$A=-B=\frac{1}{\alpha-1}\$, now we're at: $$ G(s) = \frac{1}{\alpha-1}\left(\frac{1}{s+1}-\frac{1}{s+\alpha}\right) $$
The anti transform is quite simple now since both terms in parenthesis are known transforms:
$$ g(t) = u(t)\frac{1}{\alpha-1}\left(e^{-t}-e^{-\alpha t}\right) $$
Could you get this result from your integral? Probably. Did you? Nope, and that's because integrals are so boring. Plus your book/teacher probably wanted you to use the whole A and B thing, also known as partial fraction decomposition.
The Laplace transform is a great weapon, learn how to use it and you will be rewarded.