Is there an eloquent way of qualitatively explaining why the time coordinate of an event is shifted proportionally to its spatial coordinates?

I ll take $(c=1)$ from now on.

I am not sure this is the explanation that you are looking for, but I 'll give it a try.

So for a moment let us forget about the $(x,t)$ coordinate system and focus on the normal cartesian coordinate system (i.e $(x,y)$).

Let us choose a point such that $P(x_p, y_p)$. Let us set another coordinate $(x',y')$ system such that it's rotated with $theta$ degree with respect to the x-axis. If we wanted to write the point P in terms of $(x',y')$ we need to apply a coordinate transformation, which has a form

$$begin{pmatrix} x' y' end{pmatrix}=begin{pmatrix} cos(theta) & -sin(theta) sin(theta) & cos(theta) end{pmatrix}begin{pmatrix} x y end{pmatrix} tag{1}$$

From here as you can see the length of the line element ($ds$) does not change under this transformation

$$ds^2 = dx^2+dy^2 = dx'^2+dy'^2$$

Now let us think of the SR case. In the SR the line element is defined such that

$$ds^2 = -dt^2+dx^2 = -dt'^2+dx'^2$$ (Which as you know comes from the fact that speed of light must be equal to the every inertial observer)

At this point we can naturally think a way to obtain coordinate transformation from $(x,t)$ to $(x',t')$.

Remember that $cosh^2w - sinh^2w =1$. Let me write this coordinate transformation directly

$$begin{pmatrix} x' t' end{pmatrix}=begin{pmatrix} cosh(w) & -sinh(w) -sinh(w) & cosh(w) end{pmatrix}begin{pmatrix} x t end{pmatrix} tag{2}$$

This implies that $$x' = xcosh(w) - tsinh(w)t' = -xsinh(w) + tcosh(w)$$

By using (2) you can write that

$$ds^2 = -t^2+x^2 = -t'^2 + x'^2 tag{3}$$

When you set $x=vt$ for $x'=0$ (and by using $sinh(w) = sqrt{cosh^2(w)-1}$), you will see that $cosh(w) = gamma = frac{1}{sqrt{1-v^2}}$ and $sinh(w) = vgamma$

Note: Here $w$ is not an usual angle. Its an hyperbolic angle and called rapidity

Essentially, why does time shift as a function of space just as space shifts as a function of time?

The simple answer is to preserve the $ds$ term invariant under the coordinate transformation (i.e equation 3) which is a crucial point in SR. This coordinate transformation (2) is similar to the one which I did for equation (1). So the $vx$ term comes naturally, without it the transformations do not make sense.

Edit :

So as I said we need a transformation from $(x',t')$ to $(x,t)$ or vice versa. So assume a linear form of transformation such that

$$x' = ax + bt$$ $$t' = cx + dt$$

When you impose certain conditions such that

(i) $x = vt$ for $x' = 0$

(ii) $x' = -vt'$ for $x = 0$

(iii) $x = ct$ and $x'=ct'$ (Invariance of the speed of light)

You ll find that $a = gamma$, $b = -vgamma$ etc.

I am not sure how the $cosh(w)$ and $sinh(w)$ terms are derived but It's not hard to guess.

Again take the form of equations,

$$x' = ax + bt$$ $$t' = cx + dt$$

Square them

$$x'^2 = a^2x^2 + 2axbt + b^2t^2$$ $$t'^2 = c^2x^2 + 2cxdt + d^2t^2$$

So we know that,

$x'^2 - t'^2 = x^2-t^2$

This implies

$$[a^2x^2 + 2axbt + b^2t^2] - [c^2x^2 + 2cxdt + d^2t^2] = x^2 - t^2$$

so we have $a^2-c^2 = 1$ , $ab = cd$ and $b^2 - d^2 = 1$

We also know that $cosh^2(w) - sinh^2(w) = 1$ as you can see there's a similarity between these equations so we can set

$a = b = cosh(w)$ and $c = d = sinh(w)$ to satisfy the equations.

It may not seem certain, which is not, but you can see that the form must be somehow related to the $cosh(w)$ and $sinh(w)$.

Let's say an alarm clock going off is an event.

That event shifts in space when the alarm clock is moved in space.

If the moving of the clock does not cause time dilation of the clock, then the event of the clock going off does not shift in time, when the clock is shifted in space.

Those observers that observe the clock to become more time dilated, or less time dilated, when it is being moved, agree that doubling the time that the clock is being moved doubles the amount that the event gets shifted in time because of it being shifted in space.

And doubling the time that the clock is being moved also doubles the amount that the clock/event is shifted in space.

I think I should explain what clock being moved means. Well, think about an alarm clock at the rear of a long spaceship. An astronaut slowly carries the clock to the front of the spaceship. That is what clock being moved means here. A slow change of x-coordinate in the coordinate system attached to the spaceship. An astronaut carrying the clock very slowly causes a very small amount of time dilation in the spaceship frame, but has large effect on the clock in some other frames.