$(\lambda X+\mu Y)(a)=\lambda X(a)+\mu Y(a)$ The expected value of a random variable $X$ on a sample space $\omega$ is: 
$$\begin{align*}E(X) =\text{  }\sum _{a\in \omega } X(a) \cdot P(a).\end{align*}$$

On a probability space, the expected value is linear, meaning 
that for all random variables $X$ and $Y$ and all real numbers $\lambda$ and $\mu$, 
we have 
$$\begin{align*}E(\lambda  X + \mu  Y) = \lambda  E(X) + \mu  E(Y).\end{align*}$$
Proof: This is a straightforward calculation from the definition of expected value. We have 
$$\begin{align*}E(\lambda  X + \mu  Y) =\sum _{a\in \omega }  (\lambda  X+\mu  Y)(a)\cdot P(a)=\sum _{a\in \omega }  (\lambda  X(a)+\mu  Y(a))\cdot  P(a)\end{align*}$$
Can one explain me a bit about the meaning of:
$$\begin{align*}(\lambda  X+\mu  Y)(a)=\lambda  X(a)+\mu  Y(a)\end{align*}$$
 A: For any two functions $f$ and $g$ defined on some common set $X$ with values in $\mathbb{R}$, the sum of $f$ and $g$ is defined as 
$$
(f+g)(x):=f(x)+g(x),\quad x\in X.\tag{1}
$$
Furthermore, if $a\in\mathbb{R}$ then the function multiplied by $a$ is defined as
$$
(a\cdot f)(x):=a\cdot f(x),\quad x\in X.\tag{2}
$$
Now, recall that the random variables $X$ and $Y$ are just functions from $\Omega$ into $\mathbb{R}$, and hence $(\lambda X+\mu Y)(a)$ is the function $\lambda X+\mu Y$ evaluated at $a$ which equals the pointwise evaluation $\lambda X(a)+\mu Y(a)$ using $(1)$ and $(2)$.
A: Despite its name, in probability theory, "random variable" has nothing to do with randomness (strictly speaking, "randomness" is a philosophical rather than mathemetical concept; there is no such thing as "randomness" in probability theory). A random variable $X$ is merely an outcome function that maps each point $a$ in the sample space to an outcome $X(a)$. For instance, if the sample space consists of the results of casting two dice, and $Z$ is the sum of two dice, then $\omega=\{(1,1),(1,2),\ldots,(6,6)\}$ and $Z:\omega\to\mathbb{R}$ is a function such that $Z(1,1)=2,\ Z(1,2)=3,\,\ldots,\ Z(6,6)=12$.
You may think of a random variable as an arbitrary outcome function (actually a random variable is a measurable outcome function; if you haven't learnt measure theory, ignore the meaning of "measurable" for the moment). For instances, if $\omega$ is the set of outcomes of throwing a die, i.e. $\omega=\{1,2,3,4,5,6\}$, it is perfectly fine (although it doesn't make much sense) to define two random variables as $X(a)=\sin(2\pi a)$ and $Y(a)=a^2$.
Since $X$ and $Y$ are functions, we can define a linear combination of them as usual. That is, if we denote $\lambda X+\mu Y$ by $W$, then $W$ is an outcome function (i.e. a "random variable") defined by $W(a) = (\lambda X+\mu Y)(a) := \lambda X(a)+\mu Y(a)$ for all $a\in\omega$. So, in the previous example, if $\lambda=1$ and $\mu=-3$, we have $W(a)=\sin(2\pi a)-3a^2$ for $a=1,2,\ldots,6$. If you can accept something like $\int (\color{red}{\lambda f+\mu g})=\lambda\int f+\mu\int g$ or $(\color{red}{\lambda f+\mu g})'=\lambda f'+\mu g'$ in calculus, I don't see why you have trouble to understand the meaning of $\color{red}{\lambda X+\mu Y}$ here.
A: It's a very common abuse of notation (so common that it became definition).
Generally, it is called a pointwise addition/multiplication/whatever your operator is, and is defined as
$$(f \diamond g)(\omega) \stackrel{\text{def}}{=} f(\omega) \diamond g(\omega).$$
It is worth noting, that when the domain is well-defined and there is no ambiguity (like in your case with $\Omega$), this is sometimes extended over numbers, i.e. one $ \mathbf{1} : \Omega \to \mathbb{R}$ is a constant function $\mathbf{1}(\omega) = 1$. Then you can do things like $g = (\mathbf{1}+f)$, that is $g(\omega) = (\mathbf{1}+f)(\omega) = 1 + f(\omega)$. One very common usage of this in probability theory is $Y = X - \mathbb{E}(X)$; observe that $X$ and $Y$ are functions, but $\mathbb{E}(X)$ is just a real number. Nice feature is that the multiplication by a scalar becomes the pointwise multiplication.
Note, that this is very similar to vector arithmetic:
$$(a_1,b_1,c_1) + (a_2,b_2,c_2) = (a_1 + a_2, b_1+b_2, c_1+c_2)$$
and multiplication by a scalar
$$(\lambda \cdot f)(\omega) = \lambda \cdot f(\omega)$$
is just like
$$\lambda \cdot (a,b,c) = (\lambda \cdot a, \lambda \cdot b, \lambda \cdot c).$$
It's not unusual to write $\mathbf{0}$ (or $\vec{0}$ or $\bar{0}$ or whatever) to denote a vector consisting of zeros $(0,0,\ldots,0)$. In fact vectors of length $k$ are functions $\{1,2,\ldots,k\} \to \mathbb{R}$ and functions over more complex domains can be though-of as vectors with infinitely (possibly uncountably) many coordinates.
I hope this helps ;-)
