I couldn't find a sufficiently general formula for the tangent plane of a parametric surface, so I derived one. The formula I came up with also defines the tangent line of a parametric curve in $n$-dimensions, the tangent plane to an explicitly defined surface, and the tangent plane to a parametric surface in $n$-dimensions:
Let $\mathbf{f}:\Bbb{R}^m\to\Bbb{R}^n:m<n$ be a parametrisation of an $m$-dimensional geometric object (possibly but not necessarily a manifold, algebraic variety, etc.) in Euclidean $n$-space. The parametrisation of the tangent space $T_\mathbf{u}\mathbf{f}:\Bbb{R}^m\to\Bbb{R}^n$ at a point $\mathbf{u}\in\Bbb{R}^m$ is given by...
$$T_\mathbf{u}\mathbf{f}(\mathbf{x})=\mathbf{J}_\mathbf{f}(\mathbf{u})(\mathbf{x}-\mathbf{u})+\mathbf{f}(\mathbf{u})$$
...where $\mathbf{J}_\mathbf{f}(\mathbf{u})$ is the Jacobian matrix of $\mathbf{f}$ evaluated at $\mathbf{u}$.
The resulting parametrisation preserves information about local scaling and orientation within the coordinate system. This is particularly noticeable when considering changes to the basis vectors$^*$ of the tangent space between two points $\mathbf{u}$ and $\mathbf{u}'$ along a curve, as shown below (I apologise for the quality, upload limits and all that)...
Edit
Originally I had written this formula as...
$$T_\mathbf{u}f(\mathbf{x})=\sum_i\left(\nabla f_i(\mathbf{u})\cdot(\mathbf{x}-\mathbf{u})+f_i(\mathbf{u})\right)\mathbf{e}^i$$
After Ted Shifrin's comment I realise that this is easier expressed in terms of the Jacobian. Shortly after editing this question, I found this on the Wikipedia article about the Jacobian...
$${\displaystyle \mathbf {f} (\mathbf {x} )-\mathbf {f} (\mathbf {p} )=\mathbf {J} _{\mathbf {f} }(\mathbf {p} )(\mathbf {x} -\mathbf {p})+o(\|\mathbf {x} -\mathbf {p} \|)\quad ({\text{as }}\mathbf {x} \to \mathbf {p} )}$$
...which is very close to my formula. The fact that the Jacobian is the closest linear approximation of vector-valued function explains why $T_\mathbf{u}\mathbf{f}$ is a close linear approximation of $\mathbf{f}$ near $\mathbf{u}$, but does not explain why this gives the tangent space whenever $m<n$.
Intuitively, this seems like a generalisation of Taylor's theorem to vector-valued functions. Indeed, the equation of the tangent line is given by the Taylor polynomial of degree $1$...
$$y=f'(a)(x-a)+f(a)$$
I'm not quite sure how to get from here to my formula though, or how to show that $T_\mathbf{u}\mathbf{f}$ is the tangent space of $\mathbf{f}$ for all [differentiable] $f:\Bbb{R}^m\to\Bbb{R}^n$.