The suspension (topology) and elementary examples Let $\Sigma$ denotes a suspension 
$$\Sigma X =S^1 \wedge X \equiv (S^1\times X)/(X\vee S^1)$$ 
where $\wedge$ is the the smash product, and $\vee$ is the wedge sum (one point union) of pointed topological spaces respectively.
I understand the wedge sum  $\vee$ is a "one-point union" of a family of topological spaces. But the suspension obtained by $S^1 \wedge X \equiv (S^1\times X)/(X\vee S^1)$ is harder to imagine.
Let us take elementary examples.


*

*Using the idea and the picture in Wikipedia, imagine the blue circle as $S^n$ and intuitively obtained by stretching $X=S^n$ into a cylinder and then collapsing both end faces to points.

it is easy to see that left hand side (LHS) $$\Sigma S^n=S^{n+1}$$

How do we see from the right hand side (RHS) 
  $$(S^1\times X)/(X\vee S^1)=(S^1\times S^n)/(S^n\vee S^1)=S^{n+1}?$$

Some intuition about quotient space over $(S^n\vee S^1)$ is needed.


*The suspension of a manifold is almost never a manifold. Now let us consider 
$$\Sigma T^n,$$
from the LHS I intuitively obtain the suspension $\Sigma T^n$ by stretching $X=T^n$ into a cylinder and then collapsing both end faces to points.




How do we see from the RHS 
  $$(S^1\times T^n)/(T^n\vee S^1)=(S^1\times T^n)/(T^n\vee S^1)=?$$




  
*The suspension of a manifold is almost never a manifold. Now consider the last two examples, $X=T^3 \times S^2$ and $X=T^n \times S^m$,
  what do we get on the LHS by stretching $X$ into a cylinder and then collapsing both end faces to points.
  $$\Sigma X =?$$
  what do we get on the RHS
  $$S^1 \wedge X \equiv (S^1\times X)/(X\vee S^1)=?$$ 
  How do we see that LHS and RHS match intuitively? 
  

 A: We must not confuse the (unreduced) suspension of a topological space $X$ and the reduced suspension of a pointed topological space $(X,x_0)$. Your $\Sigma X$ is the reduced suspension and should be written more precisely as $\Sigma (X,x_0)$.
Let us denote by $S X =  X \times [-1,1]/ \sim$ the unreduced suspension, where $(x,-1) \sim (x',-1)$ and $(x,1) \sim (x',1)$ for all $x,x' \in X$, and let $p : X \times [-1,1] \to SX$ denote the quotient map. Usually one defines
$$\Sigma (X,x_0) =  X \times [-1,1]/(X \times \{ -1, 1\} \cup \{ x_0 \} \times [-1,1]) = SX / p(\{ x_0 \} \times [-1,1]) .$$
We can identify $S^1$ with the quotient space $[-1,1]/\sim$, where $-1 \sim 1$. Let $\pi : [-1,1] \to S^1$ denote the quotient map and $\ast = \pi(-1) = \pi(1)$. Define
$$q :  X \times [-1,1] \to S^1 \wedge X, q(x,t) = [\pi(t),x]$$
where $[-]$ denotes equivalence class in the quotient $S^1 \times X / S^1 \vee X$. It can be shown that $q$ induces a homeomorphism $Q : \Sigma (X,x_0) \to S^1 \wedge X$.
You see that $S S^n = S^{n+1}$. $\Sigma (S^n,\ast)$ is obtained from $S^{n+1}$ by identifying half of a great circle to a point. One can show that this space is homeomorphic to $S^{n+1}$.
Questions 2 and 3 are not precise. I guess $T^n$ is the $n$-dimensional torus. Do you want to have an identification with a "well-known" space? Up to homotopy equivalence an answer is provided by Proposition 4.74 in 
Hatcher, Allen. Algebraic topology.
