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 this 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 $ΣX$.

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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.

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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.