How can output sequence be equal to the sum of copies of the impulse response, scaled and time-shifted signals?
One interpretation of your question could be as follows:


  Given that a system has the following two properties:
  
  
  the scaling or homogeneity property that if the response to input $x(t)$ is output $y(t)$, then for any choice of $\alpha$, the system response
  to scaled input $\alpha\cdot x(t)$ is scaled output $\alpha\cdot y(t)$,
  the time-invariance property that for all choices of $\tau$,
  the response to time-delayed 
  input $x(t-\tau)$ is time-delayed output $y(t-\tau)$,
  
  
  then why does the system have the additivity or superposition property
  that the response to input $x_1(t)+x_2(t)$ is 
  $y_1(t) + y_2(t)$ where
  the system response to $x_i(t)$ is $y_i(t)$, $i = 1,2~$ ???? $~~~~~~~~~~~$
  More generally, why is the system response to input 
  $\alpha\cdot x_1(t-\tau_1) +  \beta\cdot x_2(t-\tau_2)$ 
  given by $\alpha\cdot y_1(t-\tau_1) +  \beta\cdot y_2(t-\tau_2)~$?


The answer is that a system with properties 1 and 2 does not
necessarily have the additivity or superposition property.
If the superposition property also holds, then the system
is called a linear time-invariant system. But this is an
additional assumption that you need to make (or prove).

Commonly, homogeneity and additivity are combined together
into the linearity property that says that the response
to input $\alpha\cdot x_1(t)+\beta\cdot x_2(t)$ (that is,
a linear 
combination of inputs $x_1(t)$ and $x_2(t)$) is 
$\alpha\cdot y_1(t) + \beta\cdot y_2(t)$
(that is, the same linear combination of outputs $y_1(t)$ and $y_2(t)$).

A couple of points that should be tucked away into the
back of one's mind:


A system can be linear without being time-invariant (e.g.
a modulator $x(t) \to x(t)\cos(\omega t)$, or
time-invariant without being linear (e.g. a square-law
circuit $x(t) \to [x(t)]^2$
A additive system which produces output $y(t) + y(t) = 2y(t)$
in response to input $x(t) + x(t) = 2x(t)$ and so seems
to have the scaling property does not in fact have the
scaling property. Persuade yourself that this is true by
attempting to prove that the response to $0.5x(t)$ is
$0.5y(t)$. In short, scaling and additivity are two
different properties and a system that enjoys one of them
does not necessarily enjoy the other.






A second interpretation of your question could be as follows:


  For a linear time-invariant system, the output is supposed
  to be the sum of scaled and time-delayed versions of the
  impulse response, but I don't see how this is so. For example,
  the standard convolution result (for discrete-time systems)
  says
  $$y[n] = \sum_m x[m]h[n-m]$$
  where $h[\cdot]$ is the impulse (or unit) response of the
  system. But this seems to be completely backwards since the
  impulse response is running backwards in time (as in $-m$
  in the argument of $h$ in the above formula
  compared to $x[m]$ in which time is running forwards.


This is indeed a legitimate concern, but actually the
convolution formula is very successful in concealing
the result that the output is the sum of scaled and time-delayed
versions of the impulse response. What's going on is as follows.

We break down the input signal $x$ into a sum of scaled
unit pulse signals. The system response to the unit pulse signal
$\cdots, ~0, ~0, ~1, ~0, ~0, \cdots$ is
the impulse response or pulse response 
$$h[0], ~h[1], \cdots, ~h[n], \cdots$$
and so by the scaling property the single input value $x[0]$,
or, if you prefer
$$x[0](\cdots, ~0, ~0, ~1, ~0,~ 0, \cdots)
=  \cdots ~0, ~0, ~x[0], ~0, ~0, \cdots$$
creates a response
$$x[0]h[0], ~~x[0]h[1], \cdots, ~~x[0]h[n], \cdots$$

Similarly, the single input value $x[1]$ or creates
$$x[1](\cdots, ~0, ~0, ~0, ~1,~ 0, \cdots)
=  \cdots ~0, ~0, ~0, ~x[1], ~0, \cdots$$
creates a response
$$0, x[1]h[0], ~~x[1]h[1], \cdots, ~~x[1]h[n-1], x[1]h[n] \cdots$$
Notice the delay in the response to $x[1]$.  We can continue further
in this vein, but it is best to switch to a more tabular form
and show the various outputs aligned properly in time. We have
$$\begin{array}{l|l|l|l|l|l|l|l}
\text{time} \to &amp; 0 &amp;1 &amp;2 &amp; \cdots &amp; n &amp; n+1 &amp; \cdots \\
\hline
x[0] &amp; x[0]h[0] &amp;x[0]h[1] &amp;x[0]h[2] &amp; \cdots &amp;x[0]h[n] &amp; x[0]h[n+1] &amp; \cdots\\
\hline
x[1] &amp; 0 &amp; x[1]h[0] &amp;x[1]h[1] &amp; \cdots &amp;x[1]h[n-1] &amp; x[1]h[n] &amp; \cdots\\
\hline
x[2] &amp; 0 &amp; 0 &amp;x[2]h[0] &amp; \cdots &amp;x[2]h[n-2] &amp; x[2]h[n-1] &amp; \cdots\\
\hline
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \\
\hline
x[m] &amp; 0 &amp;0 &amp; 0 &amp; \cdots &amp; x[m]h[n-m] &amp; x[m]h[n-m+1] &amp; \cdots \\
\hline 
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots  
\end{array}$$
The rows in the above array are precisely the scaled and
delayed versions of the impulse response that add up to
the response $y$ to input signal $x$.
But if you ask a more specific question such as


  What is the output at time $n$?


then you can get the answer by summing the $n$-th column to get
$$y[n] = x[0]h[n] + x[1]h[n-1] + x[2]h[n-2] + \cdots + x[m]h[n-m] + \cdots
= \sum_{m=0}^{\infty} x[m]h[n-m],$$
the beloved convolution formula that befuddles generations of students
because the impulse response seems to be running backwards in time.