Let's start at the beginning:
The setting for relativity - be it special or general - is that spacetime is a manifold $\mathcal{M}$, i.e. something that is locally homeomorphic to Cartesian space $\mathbb{R}^n$ ($n = 4$ in the case of relativity), but not globally.
Such manifolds possess a tangent space $T_p\mathcal{M}$ at every point, which is where the vectors one usually talks about live. If you choose coordinates $x^i$ on the manifold, then the space of tangent vectors is
$$T_p\mathcal{M} := \{\sum_{i=0}^3 c^i \frac{\partial}{\partial x^i} \lvert c^i \in \mathbb{R} \}$$
When we say that a tupel $(c^0,c^1,c^2,c^3)$ is a vector, we mean that is corresponds to the object $c^i\partial_i \in T_p\mathcal{M}$ at some point $p \in \mathcal{M}$.
A metric on $\mathcal{M}$ can be given by specifying a non-degenerate, bilinear form at each point
$$g_p : T_p\mathcal{M} \times T_p\mathcal{M} \rightarrow \mathbb{R}$$
What you learned "in general" is that the components of the metric are, for chosen basis vectors $\partial_i$ of $T_p\mathcal{M}$, defined by $g_{ij} = g(\partial_i,\partial_j)$. You can now indeed see the metric as a kind of scalar product, setting $X \cdot Y := g(X,Y)$ for two vectors $X,Y$. (This contains the answer to your second problem) But for non-Riemannian manifolds, i.e. manifolds where not all entries in the metric are positive, this is not a scalar product in the sense you may be used to. In particular, it can be zero. Vectors for which it is zero are usually called lightlike or null.
The important thing to take away is that manifolds do not always behave like cartesian space.
Now, for your third problem, we need the concept of the cotangent space $T_p^*\mathcal{M}$. It is the dual vector space to the tangent space, spanned by the differentials $\mathrm{d}x^i : T_p\mathcal{M} \rightarrow \mathbb{R}$ for a chosen coordinate system, and defined by
$$\mathrm{d}x^i(\partial_j) = \delta^i_j$$
Now, recall that the metric was a map from twice the tangent space to $\mathbb{R}$. As such, we can see it as an element of the tensor product $T_p^*\mathcal{M} \otimes T_p^*\mathcal{M}$, which is the space spanned by element of the form $\mathcal{d}x^i \otimes \mathcal{d}x^j$. As the metric is an element of this space, it is expandable in its basis:
$$ g = g_{ij}\mathrm{d}x^i\mathrm{d}x^j$$
where the physicist just drops the bothersome $\otimes$ sign. Now, what has this to do with infinitesimal distance? We simply define the length of a path $\gamma : [a,b] \rightarrow \mathcal{M}$ to be (with $\gamma'(t)$ denoting the tangent vector to the path)$[1]$
$$ L[\gamma] := \int_a^b \sqrt{\lvert g(\gamma'(t),\gamma'(t))\rvert}\mathrm{d}t$$
And, by using physicists' sloppy notation, $g(\gamma'(t),\gamma'(t)) = g_{ij} \frac{\mathrm{d}x^i}{\mathrm{d}t}\frac{\mathrm{d}x^j}{\mathrm{d}t}$, if we understand $x^i(t)$ as the $i$-th coordinate of the point $\gamma(t)$, and so:
$$ L[\gamma] = \int_a^b \sqrt{g_{ij} \frac{\mathrm{d}x^i}{\mathrm{d}t}\frac{\mathrm{d}x^j}{\mathrm{d}t}}\mathrm{d}t = \int_a^b \sqrt{g_{ij}\mathrm{d}x^i\mathrm{d}x^j}\frac{\mathrm{d}t}{\mathrm{d}t} = \int_a^b \sqrt{g_{ij}\mathrm{d}x^i\mathrm{d}x^j}$$
Since we call $\mathrm{d}s$ the infinitesimal line element that fulfills $L = \int \mathrm{d}s$, this is suggestive of the notation
$$ \mathrm{d}s^2 = g_{ij}\mathrm{d}x^i\mathrm{d}x^j$$
If we notice that, by the definition of tangent and cotangent vectors by differentials and deriviatives as above, things with upper indices transform exactly in the opposite way from the things with lower indices (see also my answer here), it is seen that this is indeed invariant under arbitrary coordinate transformations.
$[1]$ $\gamma'(t)$ is really a tangent vector in the following sense:
Let $x : \mathcal{M} \rightarrow \mathbb{R}^n$ be a coordinate chart. Consider then: $ x \circ \gamma : [a,b] \rightarrow \mathbb{R}^n$. Since it is an ordinary function between (subsets of) cartesian spaces, it has a derivative
$$(x \circ \gamma)' : [a,b] \rightarrow \mathbb{R}^n$$
Now, $(x \circ \gamma)'^i(t)$ be be thought of as the components of the tangent vector $\gamma'(t) := (x \circ \gamma)'^i(t)\partial_i \in T_{\gamma(t)}\mathcal{M}$. It is a somewhat tedious, but worthwhile excercise to show that this definition of $\gamma'(t)$ is independent of the choice of coordinates $x$.
You exam question with the surfaces is asking about something different. You are given an embedding of a lower-dimensional submanifold $\mathcal{N}$ into Cartesian space
$$ \sigma: \mathcal{N} \hookrightarrow \mathbb{R}^n $$
and asked to calculate the induced metric on the submanifold from the Cartesian metric
$$\mathrm{d}s^2 = \sum_{i = 1}^n \mathrm{d}(x^i)^2$$
(which is just the identity matrix in component form w.r.t. any orthonormal basis of coordinates in $\mathbb{R}^n$, i.e. the dot product)
Now, how is a metric induced? Let $y : \mathbb{R}^m \rightarrow \mathcal{N}$ be coordinates for the submanifold (you are actually given $\sigma \circ y$ in the question), and $x$ be the coordinates of the Cartesian space. Observe that any morphism of manifolds $\sigma$ induces a morphism of tangent spaces
$$ \mathrm{d}\sigma_p : T_p\mathcal{N} \rightarrow T_{\sigma(p)}\mathbb{R}^n, \frac{\partial}{\partial y^i} \mapsto \sum_j \frac{\partial(\sigma \circ y)^j}{\partial y^i}\frac{\partial}{\partial x^j} $$
called the differential of $\sigma$. As a morphism of vector spaces, it is a linear map given, as a matrix, by the Jacobian $\mathrm{d}\sigma^{ij} := \frac{\partial(\sigma \circ y)^j}{\partial y^i}$ of the morphism of manifolds. Now, inducing a metric means setting
$$ g_\mathcal{N}(\frac{\partial}{\partial y^i},\frac{\partial}{\partial y^j}) := g_\mathrm{Euclidean}(\mathrm{d}\sigma(\frac{\partial}{\partial y^i}),\mathrm{d}\sigma(\frac{\partial}{\partial y^j}))$$
On the right hand side is now the dot product of two ordinary vectors in $\mathbb{R}^n$, and what your exams call $\vec e_{y^i}$ is my $\mathrm{d}\sigma(\frac{\partial}{\partial y^i})$. If you note that you are given $\sigma \circ y$, then all you need to do is to calculate the metric components by calculating $g_\mathcal{N}$ as above for every possible combination of $y^i,y^j$ (in 2D, fortunately, there's only four).
In the general case you want the Cartan-Karlhede algorithm. It is an algorithm for producing a complete set of classifying invariants for a metric, expressed as functions of the coordinates. Given the components of the metric $g$ in the coordinates $x_1, x_2, \ldots$, the algorithm produces a list
\begin{align}
\Lambda & = \Lambda(x_i) \\
\Psi_k & = \Psi_k(x_i) \quad k = 0,\ldots, 4 \\
R_{kj} & = R_{kj}(x_i) \quad k,j = 0,\ldots, 2 \\
\Lambda_{00} & = \Lambda_{00}(x_i)\\
& \;\; \vdots
\end{align}
where each quantity is defined in a way that is coordinate independent. (The names here are standard notation, but what each of them is, is a little beyond the scope of this answer.) This is in contrast to a quantity like $g_{00}$ whose value at a point depends on your coordinates. Of course, expressed as a function of coordinates, $\Lambda$ and the others may look very different in various coordinate systems, but at corresponding points, the value is the same.
Then if we have two metrics given, we can run the algorithm on both. If the metrics are really the same, but different coordinates, the invariants must agree. This gives a system of equations, \begin{align}
\Lambda(x_i) & = \Lambda'(y_j) \\
\Psi_k(x_i) & = \Psi_k'(y_j) \\
& \;\; \vdots
\end{align}
which may or may not have a solution, $x_i = x_i(y_j)$. (For example if you do this with two Schwarzschild metrics in the standard coordinates, you find that it is necessary that $m = m'$.) If there is a solution, this is your change of coordinates.
There is a caveat to the preceding. It may be that not all the equations are independent. In $n$ dimensions we need $n$ equations but the algorithm may produce fewer independent equations. This happens precisely when there is a symmetry in the spacetime. Then there cannot a unique change of coordinates, because at least one coordinate is superfluous. Indeed for the case of flat metrics the entire system is just $0 = 0$.
In this case the algorithm only establishes that there exists a change of coordinates, but you have to look at some other invariant information to find a coordinate change. (You will not be able to find a unique change of coordinates because there are many.) One piece of such information is the Killing vectors.
(This particular case is amenable to the brute force method demonstrated in the other answers, but a more complicated metric in more than two dimensions is not.)
Best Answer
If you mean you want vectors pointing in the $+,-,1,2$ directions in light-cone coordinates then you actually just want $${n_+}^\mu=\begin{pmatrix}1\\0\\0\\0\end{pmatrix},{n_-}^\mu=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix},{n_1}^\mu=\begin{pmatrix}0\\0\\1\\0\end{pmatrix},{n_2}^\mu=\begin{pmatrix}0\\0\\0\\1\end{pmatrix}.$$
(Note that the superscript $\mu$ denotes contravariant components.) A list of coordinates with a one in the $\nu$-th coordinate and zeroes elsewhere by definition denotes the vector pointing in the $\nu$-th direction. If you change the coordinate axes without changing the vector coordinates then those coordinates generally denote a different vector. Hence just declaring that you are using light-cone coordinates and not Cartesian ones is all you need to do to "find" the vectors pointing along the light-cone coordinate axes.
The things that you found are vectors pointing along the Cartesian axes, as expressed in light-cone coordinates. That's how you found them, isn't it? You took the coordinates of the Cartesian basis and put them through the change of coordinates formula. The whole purpose of the formula is to stop the physical vectors from changing by finding their new coordinates. So relabel the objects you've found as so
$${n_0}^\mu=\begin{pmatrix}1/\sqrt{2}\\0\\0\\1/\sqrt{2}\end{pmatrix},{n_1}^\mu=\begin{pmatrix} 0\\0\\1\\0\end{pmatrix},{n_2}^\mu=\begin{pmatrix}0\\0\\0\\1\end{pmatrix},{n_3}^\mu=\begin{pmatrix}1/\sqrt{2}\\0\\0\\-1/\sqrt{2}\end{pmatrix}.$$
because physically they are still the Cartesian basis vectors.
P.S. Do note that $n_+,n_-$ are not unit vectors in the sense of having spacetime interval $\pm1$.