This is a sort of continuation of this question.
In synthetic differential geometry (SDG), we have $Dsubset R$ comprised of the second order nilpotents. The KockLawvere axiom (KL axiom) implies that a function $Dtimes Dto R^n$ is of the form $a_0+a_1d_1+a_2d_2+a_3d_1d_2$. This is like a 2jet without square terms $f(a)+partial_xf _ad_1+partial_yf_ad_2+partial_{xy}f_ad_1d_2$.
In SDG, the infinitesimal rectangle $Dtimes D$ represents the second tangent bundle. In light of the KLaxioms I expect the classical second tangent bundle $mathrm T^2X=mathrm {TT}X$ of a $C^infty$ manifold admits the following kinematic description: elements are equivalence classes of germs of $C^infty$ maps $I^2to X$ where $I$ is an interval about zero, and we identify such germs if upon composing with any germ in $C_{X,x}^infty$ the partials and mixed partials coincide. Let us call such things “microsquares”. They formalize the “2jets without square terms” above.
If correct, this kinematic description is very geometric. For instance, it allows to define the flip on $mathrm T^2X$ by flipping the $x,y$ coordinates of $I^2$. The two maps $mathrm T^2Xrightrightarrows mathrm TX$ given by $mathrm Tpi_X,pi_{mathrm TX}$ are respectively given by restricting a microsquare to the $x$axis and the $y$axis. These fiber $mathrm T^2X$ in two different ways: the fiber of $mathrm Tpi_X$ over a kinematic tangent $dot gamma$ consists microsquare which restricts to $gamma$ on the $x$axis, and analogously for $pi_{mathrm TX}$.
The vertical lift applied to the tangent bundle gives a bundle isomorphism $mathrm T(mathrm TX/X)cong mathrm TXtimes_Xmathrm TX$ over $mathrm TX$, where the LHS is the vertical bundle of the tangent bundle, i.e the kernel of $mathrm Tpi_X$. For all vector bundles this acts by taking a kinematic tangent (to a fiber of the bundle) to its derivative (which is a vector in the fiber).
Question 1. How to geometrically interpret the vertical lift for a “vertical microsquare”? A microsquare lies in the vertical bundle if its restriction to the $x$axis is “constant”, i.e the derivative of the restriction is zero. This is like saying the associated “2jet without square terms” has $partial_xf_a=0$. What is the vertical lift doing with a microsquare that only makes sense if its restriction to the $x$axis is zero?
My question is motivated by another one about a seeming discrepancy between SDG and the classical $C^infty$ world:

In the $C^infty$ world, the vertical lift $
mathrm T(mathrm TX/X)cong mathrm TXtimes_Xmathrm TX$ is defined on any vertical microsquare. There is no further requirement for also being in the kernel of $pi _{mathrm TX}$ (restriction of a microsquare to its $y$axis), and I see no reason for these kernels to coincide.

In SDG, the Wraith axiom says that a function $Dtimes Dto R^n$ which is constant on the axes uniquely factors through the multiplication map $Dtimes Dto D$. This factorization takes such a function to a tangent vector, and this is the analog of the vertical lift. The $C^infty$ version of being constant on the axes is having the $partial_x,partial_y$ coefficients of the ‘2jet without square terms’ vanish $partial_xf_a=0=partial_yf_a$. The remaining mixed partial term indeed factors through the multiplication map because that’s how Taylor series are. The point is that the Wraith axiom asks for both partials to vanish, as opposed to the vertical bundle which involves only vanishing $partial_x$.
Question 2. What is going on here, geometrically? Why does SDG want both partials to vanish while the $C^infty$ world only cares about one of the partials?
Lastly and perhaps most fundamentally: I don’t understand the geometric meaning of a microsquare. I understand 2jets since we retain the information of the Hessian, but retaining only the mixed partials – I don’t get it.
Question 3. What is the geometric content of a microsquare / an element in the second tangent bundle?