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 $2B$.
We have $\mathrm{Tor}(A,B)=0$ if $A$ is projective or flat. So we need our projective objects to be the input of $A\otimes -,$
if we're viewing the projective objects as right $R-$modules. The tensor functor is right exact, so given any projective resolution of $B$
$$\cdots \overset{}{\longrightarrow}P_n\overset{d_n}{\longrightarrow} P_{n-1}\overset{}{\longrightarrow}\cdots \overset{d_1}{\longrightarrow}P_0\overset{\alpha}{\longrightarrow}B\overset{}{\longrightarrow}0$$
then the tensor functor $A\otimes-$
$$\cdots \overset{}{\longrightarrow}A\otimes P_n\overset{1\otimes d_n}{\longrightarrow} A\otimes P_{n-1}\overset{}{\longrightarrow}\cdots \overset{1\otimes d_1}{\longrightarrow}A\otimes P_0\overset{1\otimes\alpha}{\longrightarrow}A\otimes B\overset{}{\longrightarrow}0$$
is a chain complex, and $$\mathrm{Tor}(A,B)=\mathrm{ker}(1\otimes d_n)~/~\mathrm{image}(1\otimes d_{n+1}).$$
I don't know of any way to define $\mathrm{Tor}$ using injective resolution. [$\S17.1$ Dummit and Foote]
Best Answer
If $A$ and $B$ are objects in the abelian category $\mathcal{A}$, the (equivalence class) of the exact sequence $$0\to A\to A\oplus B\to B\to 0$$ is the zero object in $\text{Ext}^{1}(B,A)$.
For $n>1$, the zero object of $\text{Ext}^{n}(B,A)$ is the (equivalence class) of the sequence
$$0 \to A\to A \to 0 \to\cdots\to 0\to B\to B\to 0$$ where the maps $A\to A$ and $B\to B$ are the identity.
The group structure on both is a bit fiddly. Consider the maps $\Delta:A\to A\oplus A$ given by $\begin{pmatrix}1 \\ 1\end{pmatrix}$ and $\nabla:A\oplus A\to A$ given by $\begin{pmatrix}1 & 1\end{pmatrix}$. The operation $+$ on $\text{Ext}^{i}(B,A)$ is $$\textbf{E}_{1}+\textbf{E}_{2} := \nabla (\textbf{E}_{1}\oplus\textbf{E}_{2})\Delta$$ where the direct sum is what you expect.
A proof that these form an abelian group can be found in Chapter 7 (called Extensions) of book Theory of Categories by Mitchell. It also shows that these coincide with the usual derived functor definition of Ext groups.