Gradient of parametric surface

The function $\vec{V}$ gives the surface parametrically as:

$$ \vec{V}(s,t) = \left< x(s,t), \, y(s,t), \, z(s,t) \right> $$

You have the partial derivative vectors:

$$ \frac{\partial \vec{V}}{\partial s} = \left< \frac{\partial x}{\partial s}, \, \frac{\partial y}{\partial s}, \, \frac{\partial z}{\partial s} \right> $$

$$ \frac{\partial \vec{V}}{\partial t} = \left< \frac{\partial x}{\partial t}, \, \frac{\partial y}{\partial t}, \, \frac{\partial z}{\partial t} \right> $$

The $\left[ \frac{\partial V}{\partial s}, \, \frac{\partial V}{\partial t}\right]$ that you suggest wouldn't really be a vector, so I'm not sure it makes sense to say it "points" in some direction. But you can interpret it as the Jacobian matrix

$$ DV = \left( \begin{array}{cc} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t} \\ \frac{\partial z}{\partial s} & \frac{\partial z}{\partial t} \end{array} \right) $$

You could interpret the rows of this matrix as the gradients $\nabla x$, $\nabla y$, and $\nabla z$. Then the last row, $\nabla z$, would point in the direction (in the $s,t$-plane) of greatest increase in $z$.

A "direction of steepest ascend" is generally not defined for parametric surfaces embedded in 3D space. Unless your parametric surface also possesses an explicit representation (which is not always possible https://youtu.be/jZRqCfi5_Uo?si=I8g161Jg1w07iHjO), there is no natural/intuitive sense of "up" direction along which you can travel in order to "ascend".

For an explicit surface $z = f(x,y)$, like the one shown above, it is understood/agreed that the direction of ascension is the same as that of the $\hat{z}$ cartesian vector. Looking at the graph above, it becomes clear why this convention appears so natural to us, as we can intuitively think of the variable $z$ as the 'height' of a 'mountain' we are trying to climb starting from its bottom.

But, for a general parametric surface $\vec{r}(u,v) = x(u,v)\hat{x} + y(u,v)\hat{y} + z(u,v)\hat{z}$, with no explicit representation, there is no natural/intuitive sense of which direction is "up". Unlike in the previous example, looking now at the graph above leaves us disoriented as to where the 'bottom' or the 'top' of the graph should be. All we see is a knot; no intuitive or generally-agreed-upon ascension direction comes to mind. The surface just exists in 3D space.

Now, if you really wanted to, what you could do is define your own ascension direction manually (agreed upon only by you) and then work from there. Something like that would work as follows:

First, let $\vec{r} = x\hat{x} + y\hat{y} + z\hat{z}$ + describe a point in real 3D-space and let $\vec{r} = u\hat{u} + v\hat{v} + w\hat{w}$ describe a point in parametric 3D-space. Also, $\hat{x}, \hat{y}, \hat{z}$ form an orthonormal basis for real 3D-space and $\hat{u}, \hat{v}, \hat{w}$ form an orthonormal basis for parametric 3D-space.

Then, for any parametric surface: $\vec{r}(u,v) = x(u,v)\hat{x} + y(u,v)\hat{y} + z(u,v)\hat{z}$ in real 3D-space, we can define an arbitrary ascension direction using a constant unit vector: $\hat{d} = a\hat{x} + b\hat{y} + c\hat{z}$. It follows then that the function $\lambda(u,v) = \hat{d} \cdot \vec{r}(u,v)$ yields the distance of any point on the surface, from the origin, along the direction $\hat{d}$. Expanding $\lambda$ produces:

$\lambda(u,v) = ax(u,v) + by(u,v) + cz(u,v)$

Which we recognize as a scalar function on the $u,v$-plane. Consequently, $w = \lambda(u,v)$ describes an explicit surface on parametric 3D-space. Computing the gradient of this surface then yields:

$\nabla{w} = \nabla{\lambda(u,v)} = \frac{\partial \lambda(u,v)}{\partial u}\hat{u} + \frac{\partial \lambda(u,v)}{\partial v}\hat{v}$

$\nabla{w} = \nabla{\lambda(u,v)} = (a\frac{\partial x(u,v)}{\partial u} + b\frac{\partial y(u,v)}{\partial u}+ c\frac{\partial z(u,v)}{\partial u})\hat{u} + (a\frac{\partial x(u,v)}{\partial v} + b\frac{\partial y(u,v)}{\partial v} + c\frac{\partial z(u,v)}{\partial v})\hat{v}$

$\nabla{w} = \nabla{\lambda(u,v)} = (\hat{d} \cdot \frac{\partial \vec{r}(u,v)}{\partial u})\hat{u} + (\hat{d} \cdot \frac{\partial \vec{r}(u,v)}{\partial v})\hat{v}$.

Thus: $(\hat{d} \cdot \frac{\partial \vec{r}(u,v)}{\partial u})\hat{u}$ and $(\hat{d} \cdot \frac{\partial \vec{r}(u,v)}{\partial v})\hat{v}$ are vectors in the input space (i.e, parametric space) which when followed generate the greatest ascent in the output space (i.e, real space), measured as the distance from the origin along direction $\hat{d}$. Note that $\frac{\partial \vec{r}}{\partial u}$ and $\frac{\partial \vec{r}}{\partial v}$ are indeed the columns of the Jacobian matrix of the surface, just as @Nick had stated.

$J = \begin{bmatrix}\frac{\partial \vec{r}}{\partial u} & \frac{\partial \vec{r}}{\partial v}\\\end{bmatrix} = \begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v}\\ \end{bmatrix}$

The answer given by @Nick was basically perfect (and way shorter). I just thought I'd expand on it a bit more and actually answer how you'd go about calculating the steepest ascend in a parametric surface, given a predefined ascent direction, if you wanted to.