It is known that in a stable $\infty$-category $\mathcal{C}$ and $X \in \mathcal{C}$, the suspension $X[1]$ is defined by the pushout of $0\leftarrow X \rightarrow 0$. However this does not make sense in usual category since the pushout is the zero object $0$.
Now all we can do is following the formal definition of limit/colimit for an $\infty$-category. But the cost is that we lose the intuition. What (intuition) should one keep in mind when thinking about pushout/pullback in a stable $\infty$-category.
Best Answer
Let me try to give some intuition by examining two important examples. One should start from the definition: the suspension $ΣX$ is the universal choice of $Y$ filling of a square $$\require{AMScd} \begin{CD} X @>{p}>> \ast\\ @V{p}VV @VVV \\ \ast @>>> Y \end{CD}\,.$$ To understand the suspension we need to understand what's the datum of such a square. This is given by two maps $y_0:\ast\to Y$ and $y_1:\ast\to Y$ (two ``points'' of $Y$) and a homotopy $H:y_0p\simeq y_1p$ as maps $X\to Y$. That is, the suspension is the universal recipient of two points and a homotopy between the two constant maps at $y_0$ and $y_1$.
Let us go now in the ∞-category of spaces (or, if you prefer the name, animæ). Then a homotopy of maps $H:X\to Y$ is exactly a map $H:X\times [0,1]\to Y$. Since we require the homotopy to be between two constant maps we see that a commuting square as above is exactly the datum of a map $$\ast\amalg_{X\times\{0\}}X\times [0,1]\amalg_{X\times\{1\}} \ast\to Y$$ One needs to check the universal property more carefully, but as one would expect the left hand side is exactly the universal recipient. That is $$ ΣX=\ast\amalg_{X\times\{0\}}X\times [0,1]\amalg_{X\times\{1\}} \ast$$ Note that the right hand side is exactly the classical suspension of $X$, thus justifying the name $ΣX$.
Now for a more algebraic (and stable) example: the derived category of a ring. In this case, when representing everything by chain complexes, the two points are no data (since there's only one possible map $0\to Y$), and a homotopy is the same as a collection of maps $H_n:X_{n+1}\to Y_n$ such that $dH_n+H_nd=0$. But this is exactly the same as a map of chain complexes $X[1]\to Y$ (remember than in $X[1]$ the differential inherits a sign to make the formulas canonical -- this is exactly its origin!). Therefore in $\mathscr{D}(R)$, the suspension is given by the usual shift.
So what's the reason for these new phenomena? The point is that a commuting square in an ∞-category in general contains more information than the square in a 1-category: it's not enough that you say that the two composition are equal, you also have to provide a reason for them to be equal (i.e. a homotopy). Therefore your pushout needs to account for this additional data, and this is why it is nontrivial.