Write down the equations of the tangent space $T_aX$ to a submanifold $X = \{f_1 = \cdots = f_k = 0 \}$ backwards $\in a = \{a_1, \cdots a_n \}$. What is the dimension of $T_aX$.
I wrote the equations to be
$$\frac{\partial f_1}{\partial x_1} (a) (x_1 - a_1) + \cdots \frac{\partial f_1}{\partial x_k}(a)(x_k - a_k) + \cdots + \frac{\partial f_k}{\partial x_1} (a) (x_1 - a_1) + \cdots \frac{\partial f_k}{\partial x_k}(a)(x_k - a_k)$$
but I don't know what the dimension is. What is it?
The dimension of the tangent space of a smooth manifold at a point is always the same as the dimension of the manifold. The dimension of the tangent space of a manifold is twice the dimension of the dimension of the manifold.
In your case, $k$ is the codimension of the manifold (as it has been pointed out in the comments). This means that if your manifold is a submanifold of $\mathbb{R}^n$ (with $n>k$), then the manifold has dimension $n-k$.
In order to see why it is so, I propose you the following two examples:
The sphere $S^2\subset\mathbb{R}^3$ is given as level set of the single function $f_1:\mathbb{R}^3\rightarrow\mathbb{R},\ (x,y,z)\mapsto x^2+y^2+z^3-1$, that is: $S^2 = \{(x,y,z)\in\mathbb{R}^3|f_1(x,y,z) = 0\}$. As you can see $\dim S^2 = 2$, and its codimension is $1$.
The circle $S^1\subset\mathbb{R}^3$ given as level set of the two functions $f_1:\mathbb{R}^3\rightarrow\mathbb{R},\ (x,y,z)\mapsto x^2+y^2-1$ and $f_2:\mathbb{R}^3\rightarrow\mathbb{R},\ (x,y,z)\mapsto z$ that is: $S^1 = \{(x,y,z)\in\mathbb{R}^3|f_1(x,y,z) = 0,\ f_2(x,y,z)=0\}$. As you can see $\dim S^1 = 1$, and its codimension is $2$.
