adding u(t) and r(t) to create a signal
I'm not sure I understand the delta-equals notation, but I'll assume it means definition.

It's important to understand that you're doing a variable substitution here.  In the strictest sense, the $t$ in $u(t-1)$ is not the same $t$ as in $u(t)$.  It may be easier for you to understand if you write your definitions as functions of $z$ instead of $t$:

$$ u(z) \triangleq
\begin{cases}
0, &amp; \text{if }z&lt;0 \\
1, &amp; \text{if }z\ge0
\end{cases} $$

$$ \begin{align}
r(z) &amp;  \triangleq
\begin{cases}
0, &amp; \text{if }z&lt;0 \\
z, &amp; \text{if }z\ge0
\end{cases} \\
&amp; = z \ u(z) \\
\end{align}$$

So, $z=t$ in $u(t)$ and $z=t-1$ in $r(t-1)$, etc.  You're changing a function of $z$ into a function of $t$.

The two functions you're dealing with have a breakpoint in them where the function changes behavior.  In both cases it happens when $z=0$.  So, looking at $u(z)$ for example, $u(z)=0$ for all values of $z$ less than zero and $u(z)=1$ for $z=0$ and all values greater than zero-- just saying what the function says, but in words.  

When I use $u(t)$, I am substituting $z=t$ so $u(t)=0$ for all times before $t=0$, and $u(t)=1$ from time $t=0$ onward.

Now think about what's happening in the function argument with all the $(t-1)$, $(t-2)$ values...  If you were to look at $u(t-4)$, then, it has the effect of shifting the entire function to the right in time.  In this case:
$$ u(t-4) =
\begin{cases}
0, &amp; \text{if }t-4&lt;0 \\
1, &amp; \text{if }t-4\ge0
\end{cases} $$
or, rearranging the conditions a bit:
$$ u(t-4) =
\begin{cases}
0, &amp; \text{if }t&lt;4 \\
1, &amp; \text{if }t\ge4
\end{cases} $$
so you have a step function that is zero for all times before $t=4$, and one for times $t=4$ onward.

With the $r(z)$ function, with $z=t-2$ you substitute for $z$ in all places, so you have:
$$ \begin{align}
r(t-2) &amp;  =
\begin{cases}
0, &amp; \text{if }t&lt;2 \\
t-2, &amp; \text{if }t\ge 2
\end{cases} \\
&amp; = (t-2) \cdot u(t-2) \\
\end{align}$$
which is zero for all times before $t=2$ and then a line starting at $(t=2,x(t)=0)$ with a slope of positive 1 and a y-intercept of -2.

The coefficients have the effect of changing the slope and corresponding y-intercept but the t-intercept remains the same.  $2\cdot r(t-2)$ has a slope of 2 and a y-intercept of -4 but still crosses the $t$ axis at $t=2$.

The x(t) function is a linear combination of such functions, so you must add up the values at each point.  The easiest way to see this on graph paper would be to draw each term separately one above the other and then add them vertically.  

With the two components you you've highlighted, $−r(t−1)+2\cdot r(t−2)$:  There is a slope of zero before $t=1$ because $u(t)$ is zero slope.  The first term starts a ramp of slope of -1 starting at $t=1$ and the second term adds to that a ramp of slope +2 starting at $t=2$.  Where those two terms overlap (everywhere from $t=2$ and to the right) the combined slope of those terms is +1.