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 $ΣX$.

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.

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.