As you said, you can compute $\operatorname{Ext}(\mathbb{Z}/2\mathbb{Z},\cdot)$ using a projective resolution of $\mathbb{Z}/2\mathbb{Z}$. But you can also use this projective resolution to understand what the maps $f,g,h$ are ! You will find out that the maps $g$ and $h$ are the obvious one, namely a map $X\rightarrow Y$ induces a map $X/2X\rightarrow Y/2Y$, and those are the ones that arises in the long sequence of $\operatorname{Ext}$.
The map $f$ is a connecting homomorphism and as such it is not always very obvious. As often it comes from the snake lemma, but here you can draw the entire diagram to get everything :
$$\require{AMScd}
\begin{CD}
0@>>>\operatorname{Hom}(\mathbb{Z}/2\mathbb{Z},B)@>>>\operatorname{Hom}(\mathbb{Z}/2\mathbb{Z},X)@>>>\operatorname{Hom}(\mathbb{Z}/2\mathbb{Z},A)\\
@.@VVV@VVV@VVV\\
0@>>>\operatorname{Hom}(\mathbb{Z},B)@>>>\operatorname{Hom}(\mathbb{Z},X)@>>>\operatorname{Hom}(\mathbb{Z},A)@>>>0\\
@.@V{\times 2}VV@V{\times 2}VV@V{\times 2}VV\\
0@>>>\operatorname{Hom}(\mathbb{Z},B)@>>>\operatorname{Hom}(\mathbb{Z},X)@>>>\operatorname{Hom}(\mathbb{Z},A)@>>>0\\
@.@VVV@VVV@VVV\\
@.\operatorname{Ext}(\mathbb{Z}/2\mathbb{Z},B)@>>>\operatorname{Ext}(\mathbb{Z}/2\mathbb{Z},X)@>>>\operatorname{Ext}(\mathbb{Z}/2\mathbb{Z},A)@>>>0
\end{CD}
$$
or using identifications $\operatorname{Hom}(\mathbb{Z},X)=X$ and $\operatorname{Hom}(\mathbb{Z}/2\mathbb{Z},X)=X[2]$, this diagram can be written :
$$\require{AMScd}
\begin{CD}
0@>>>B[2]@>>>X[2]@>>>A[2]\\
@.@VVV@VVV@VVV\\
0@>>>B@>>>X@>>>A@>>>0\\
@.@V{\times 2}VV@V{\times 2}VV@V{\times 2}VV\\
0@>>>B@>>>X@>>>A@>>>0\\
@.@VVV@VVV@VVV\\
@.B/2B@>>>X/2X@>>>A/2A@>>>0
\end{CD}
$$
with the natural maps. The connecting homomorphism from the snake lemma gives a long exact sequence
$$0\rightarrow B[2]\rightarrow X[2]\rightarrow A[2]\overset{f}\rightarrow B/2B\rightarrow X/2X\rightarrow A/2A\rightarrow 0$$
Finally the map $f$ is the following : take an element $a$ of order 2 in $A$, lift it in $X$ to get an element $x$. Multiply $x$ by 2, it will land in $B$ and take its class modulo $2B$.
Best Answer
Sometimes one calls a left exact sequence an exact sequence of the form $$\tag{*} 0 \to A' \to A \to A''$$ and a right exact sequence an exact sequence of the form $$\tag{**} A' \to A \to A'' \to 0$$ but this terminology is not very common (I would try to avoid it), and there's no such thing as a left/right exact sequence of arbitrary length, as far as I know.
A sequence of morphisms $$\cdots \to A_{n+1} \xrightarrow{f_{n+1}} A_n \xrightarrow{f_n} A_{n-1} \to \cdots$$ is exact if $\ker f_n = \operatorname{im} f_{n+1}$ for all $n$, and it is a complex if $\operatorname{im} f_{n+1} \subseteq \ker f_n$ for all $n$. Note that an exact sequence is the same thing as a complex with trivial (co)homology.
The condition $\operatorname{im} f_{n+1} \subseteq \ker f_n$ is equivalent to $f_n\circ f_{n+1} = 0$, and after applying any additive functor $F$ you get $$F (f_n)\circ F (f_{n+1}) = F (f_n\circ f_{n+1}) = F (0) = 0,$$ so that any additive functor maps a complex (in particular, an exact sequence) to a complex. Similarly for contravariant $F$, $$F (f_{n+1})\circ F (f_n) = F (f_n\circ f_{n+1}) = F (0) = 0.$$
If $F$ is exact (i.e. both left and right exact), then it preserves (co)homology: $$H_n (F (C_\bullet)) \cong F (H_n (C_\bullet)).$$ In particular, an exact functor maps an exact sequence of arbitrary length to an exact sequence. If $F$ is contravariant, $$H^n (F (C_\bullet)) \cong F (H_n (C_\bullet)).$$
If $F$ is not exact (not left/right exact), then there is no reason for $F$ to preserve (co)homology of a complex.
In particular, if you start with a resolution $P_\bullet \twoheadrightarrow A$, then even though the complex $$\cdots \to P_2 \to P_1 \to P_0 \to 0$$ is exact everywhere, except for $P_0$, the corresponding sequence of morphisms $$0 \to \operatorname{Hom} (P_0,B) \to \operatorname{Hom} (P_1,B) \to \operatorname{Hom} (P_2,B) \to \cdots$$ is still a complex, but a priori there is no reason for it to be exact anywhere. Left exactness of $\operatorname{Hom} (-,B)$ will tell you only that the exact sequence $$P_1 \to P_0 \to A \to 0$$ corresponds to the exact sequence $$0 \to \operatorname{Hom} (A,B) \to \operatorname{Hom} (P_0,B) \to \operatorname{Hom} (P_1,B)$$ that is,
\begin{multline} H^0 (\operatorname{Hom} (P_\bullet,B)) = \ker (\operatorname{Hom} (P_0,B) \to \operatorname{Hom} (P_1,B)) \cong \\ \operatorname{Hom} (\operatorname{coker} (P_1 \to P_0), B) \cong \operatorname{Hom} (A,B). \end{multline}
so that $\operatorname{Ext}^0 (A,B) \cong \operatorname{Hom} (A,B)$.
If $\operatorname{Hom} (-,B)$ is an exact functor (in this case one says that $B$ is injective), then it indeed preserves (co)homology, and $\operatorname{Ext}^n (A,B) = 0$ for all $A$ and $n > 0$.
Here's a minimal example with abelian groups: to calculate $\operatorname{Ext} (\mathbb{Z}/n\mathbb{Z}, B)$, we may start with a projective resolution of $\mathbb{Z}/n\mathbb{Z}$: $$0 \to \mathbb{Z} \xrightarrow{\times n} \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to 0$$ After applying $\operatorname{Hom} (-,B)$ to the complex $$0 \to \mathbb{Z} \xrightarrow{\times n} \mathbb{Z} \to 0$$ you get the complex $$0 \to B \xrightarrow{\times n} B \to 0$$ Its cohomology in degree $0$ is $$\operatorname{Ext}^0 (\mathbb{Z}/n\mathbb{Z}, B) = \ker (B \xrightarrow{\times n} B) = \operatorname{Hom} (\mathbb{Z}/n\mathbb{Z}, B),$$ and its cohomology in degree $1$ is $$\operatorname{Ext}^1 (\mathbb{Z}/n\mathbb{Z}, B) = \operatorname{coker} (B \xrightarrow{\times n} B) = B/nB,$$ which is not trivial if the multiplication by $n$ is not surjective on $B$.