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Let $\textbf{C}/C$ be a slice category with base object $C$.

The functor that comes to mind is the one such that objects $f:X \to C$ are mapped to $X$, and arrows $a: X \to X'$ are mapped to themselves.

Let $F$ be this mapping. Then since, the identiy on the object $f: X\to C$, is $1_X$, we have that $F(1_f) = F(1_X) = 1_{F(X)}$. Further, if $a: f \to g$, then define $F(a) = a$, for which we have $F(a): F(f) \to F(g)$, and finally $F(i \circ j) = i \circ j$.

First of all how can I make it explicit that $a : f \to g$ in $\textbf{C}/C$, but in $\textbf{C}$, $a: \text{dom}(f) \to \text{dom}(g)$?

And then how is $C$ "forgotten"?

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$a : f \to g$ is already an explicit way to say $a : f \to g$. If you want to put the category in the notation, a simple way is to use a homset; two notations for such are

$$ a \in \hom_{\mathcal{C}/C}(f, g) \qquad \qquad \text{or} \qquad \qquad a \in (\mathcal{C}/C)(f, g) $$

Alternatively, rather than writing $f$ for an object of $\mathcal{C}/C$, you may instead write $(X,f)$ for the object. Or even the whole diagram $X \xrightarrow{f} C$. Then maybe having both $a : (X,f) \to (Y,g)$ and $a : X \to Y$ is already sufficient for your tastes.

As for forgetting $C$, it's just what it sounds like; applying $F$ just discards everything related to $C$. For example, on the left is a depiction of a commutative triangle in $\mathcal{C} / C$. Applying $F$ just lops off everything crossing the dotted line, to give the triangle on the right.

enter image description here

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    What about the object $h : C \to C$ in $\textbf{C}/C$. Its image under $F$ is $C$ and $h$ is a morphism in $\textbf{C}/C$, if we take $f = 1_C, g = 1_C$? So isn't $C$ still in the image of $F$? – 2017-01-23
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    @Fruitful; Yes, any $C$'s to the left of the dotted line would still remain. I would have put something like that in my image if I had thought of it at the time, but I'm away from my software to redraw it now. – 2017-01-23
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    Okay, so it's describing that the other objects forgot about and left $C$ isolated. – 2017-01-23
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If $f:X\rightarrow C$ and $g:Y\rightarrow C$, $a:f\rightarrow g$ is a morphism $a:X\rightarrow Y$ such that $f=g\circ a$.

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All you need to do is say what $F$ does to objects and morphisms, and then check that it is a functor. An object in $C/C$ is of the form $a:X\to C$ and $F$ sends it to $X$. A morphism from $a:X\to C$ to $\ b:Y\to C$ is a morphism $\phi:X\to Y$ such that $b\circ \phi =a$. Now, $F$ simply sends $\phi$ to $\phi,\ $and forgets the commuting triangle. Of course, you need to check associativity and the fact that $F1_X=1_{FX},\ $but these are pretty obvious.

Actually, there are several different "forgetful" functors from slice categories. See here.