Definition $:$ A semi-simplicial set is a sequence of discrete space $\{X_n\}_{n \geq 0},$ where $X_n$ is a discrete space consisting of $n$-simplices for $n \geq 0$ together with the face maps $d_i : X^n \longrightarrow X^{n-1},$ for $0 \leq i \leq n$ satisfying simplicial identities $d_i d_j = d_{j-1} d_i,$ for $i \lt j.$
Given a semi-simplicial set $X$ (also known as s$\Delta$-set) we define it's geometric realization $|X|$ as follows $:$
Let $|X|^0 = X_0$ (the discrete space of $0$-simplices or points in $X$). After having $|X|^{n-1}$ define $|X|^n$ inductively as the pushout of the following diagram $:$
$$\require{AMScd} \begin{CD} X_n \times \partial \Delta^n @>{}>{}> X_n \times \Delta^n \\ @V{\Phi_n}VV @VV{}V\\ |X|^{n-1} @>{}>{\text {}}> |X|^n \end{CD}$$
Now in order to define $\Phi_n$ it is enough to define it for the sets $X_n \times \partial_i \Delta^n,$ since $\partial \Delta^n = \bigcup\limits_{i = 0}^{n} \partial_i \Delta^n.$ Given an element $\sigma \in X_n$ and $t = (t_0,\cdots,t_{i-1},0,t_{i+1}, \cdots, t_n) \in \partial \Delta^n \subseteq \Delta^{n+1}$ define $\Phi_n$ by the following composition of maps
$$X_n \times \partial_i \Delta^n \xrightarrow{(\sigma,t) \longmapsto (d_i (\sigma),\hat {t_i})} X_{n-1} \times \Delta^{n-1} \longrightarrow |X|^{n-1}$$
where $\hat {t_i} = (t_0,\cdots,t_{i-1},t_{i+1}, \cdots , t_n) \in \Delta^{n-1}$ and the second map is obtained from the pushout of $|X|^{n-1}.$
Now let $|X| : = \bigcup\limits_{n \geq 0} |X|^{n-1}$ endowed with the weak topology i.e. a subset $U \subseteq |X|$ is open if and only if $U \cap |X|^n$ is open in $|X|^n,$ for all $n \geq 0.$
Then $|X|$ is called the realization of the semi-simplicial set $X.$ Since $X_n$ is a discrete space and $\Delta^n \cong D^n$ and $\partial \Delta^n \cong S^{n-1},$ it follows that $|X|$ is a CW-complex with the set of $n$-cells being $X_n$ and the $n$-skeleton being $|X|^n.$
$\textbf{Question}$ $:$ How do I define $\Phi_1\ $? Since $|X|^{0}$ is not obtained as a pushout so I don't have the access of a map from $X_0 \times \Delta^0 \longrightarrow |X|^0 = X_0.$
Our instructor gave some examples of s$\Delta$-sets such as circle, torus etc. Circle can be given a structure of an s$\Delta$-set by means of the following diagram $:$
Here there are two $0$-simplices namely $v$ and $w$ and two $1$-simplices namely $a$ and $b.$ The face maps $d_0$ and $d_1$ are given by $d_0(a) = d_1(b) = w$ and $d_0(b) = d_1(a) = v.$ Here $d_0$ and $d_1$ satisfy the only simplicial identity $d_0d_1 = d_0^2,$ which is vacuously satisfied. So the circle has the structure of an $s$$\Delta$-set. Now since there are no $n$-simplices for $n \geq 2,$ so it is clear that $|\underline {S^1}| = |\underline {S^1}|^1,$ where $\underline {S^1}$ is the underlying s$\Delta$-set of $S^1.$ Here our instructor defined $\Phi_1 : \{a,b\} \times \{0,1\} \longrightarrow \{v,w\}$ as follows $:$
$\Phi_1 (a,0) = d_0(a) = v,$
$\Phi_1 (a,1) = d_1(a) = w,$
$\Phi_1 (b,0) = d_0 (b) = w,$
$\Phi_1 (b,1) = d_1(b) = v.$
Then $|\underline {S^1}|^1$ can be obtained from the coproduct $(\{a,b\} \times \{0,1\}) \sqcup (\{a,b\} \times [0,1])$ by the following identifications $:$
$(a,0) \sim (b,1) \sim v$ and
$(a,1) \sim (b,0) \sim w$
which can easily be seen to be circle. So the realization of a circle as an s$\Delta$-set is again a circle.
Similarly torus can also be viewed as an s$\Delta$-set as follows $:$ 
Here there is only one $0$-simplex namely $v,$ three one $1$-simplices namely $a,b$ and $c$ and two $2$-simplices namely $\Delta_1$ and $\Delta_2.$ The face maps on $1$-simplices send all of $a,b$ and $c$ to $v$ and the face maps $d_0,d_1$ and $d_2$ on $2$-simplices are given as follows $:$
$d_0(\Delta_1) = a, d_0(\Delta_2) = b.$
$d_1(\Delta_1) = c, d_1(\Delta_2) = c.$
$d_2(\Delta_1) = b, d_2(\Delta_2) = a.$
Actually here $d_i (\sigma) =$ face opposite to the $i$-th vertex of $\sigma.$ Then clearly $d_i$'s will satisfy simplicial identities since there is only one $0$-simplex $v$ in this case. So torus has a structure of an s$\Delta$-set. Now I am willing to find out the realization of the torus.
If we take the same definition of $\Phi_1$ as in the case of circle I find that $|\underline {T}|^1$ is the wedge of three circles pinched at $v.$ I am not quite sure whether it is indeed the case or not. But I got stuck in finding $|\underline {T}|^2$ which is the required realization of the torus as an s$\Delta$-set. Here I failed to find $\Phi_2.$
Would anybody please help me in this regard? As I am a beginner of this subject I need some help in this regard.
Thanks for reading.
Note that, $\Phi_1$ is a map from $X_1\times \Delta^1\longrightarrow |X|^0 = X_0$ ,so your question doesn't make any sense. Next, your answer is correct for $|T^1|$ it's $S^1\vee S^1\vee S^1$ which is quite easy to see. Now, we assigned a direction to each of this $3$ loops $a,b,c$ and we will think them as directed path.
We think $T^2\times \Delta^2$ as square that you have drawn, except we don't identify the opposite edges and the diagonal $c$ think them as disjoint two triangles with an ordered vertices , call first triangle $\Delta_A$ and second one as $\Delta_B$
Now, we define, $\Phi_2$ as follows, first consider a map $\Delta_A\times \partial_0\Delta^2\rightarrow (a,(t_1,t_2))$ where $(a,(t_1,t_2))$ correspond to the edge $[1,2]$ of $\Delta_A$ and then a map from $(a,(t_1,t_2)$ to the loop $a$ according to its direction and similarly for others.