There are a lot of nice theorems about flat modules over general rings. For example, over a domain, any flat module is torsion-free, and the converse holds over a principal ideal domain. (Note that the notion of being torsion-free does not make sense over rings that are not domains). Qiaochu mentioned some other nice theorems. Often, there is a bit of interplay between the algebra and the geometry, and it depends on your situation which approach you want to use.
But the fibre dimension criterion is not as useless as you think! For example, consider the example that you give of $k[t^2,t^3] \subseteq k[t]$ (I assume this is the example you mean; for this you have adjoin $\frac{y}{x}$, not $\frac{x}{y}$, because $x$ corresponds to $t^2$ and $y$ to $t^3$). Then you probably know that this gives an isomorphism away from the origin, i.e. when we invert $t$ it becomes an isomorphism. Thus, over all primes not equal to $(t^2,t^3)$, the fibre dimension is $1$.
However, at the prime $(t^2,t^3)$, we can just compute:
$$\dim_k k[t]/(t^2,t^3) = 2,$$
with basis given by $1, t$ (note that for a maximal ideal $\mathfrak m$, we do need to localise, because $A/\mathfrak m$ is already a field). Thus, this extension of rings is not flat.
But I wouldn't say that there is a general method for testing whether an extension of rings is flat. In geometric situations (i.e. where both rings are finite type algebras over a field), often the best way to decide such questions is by looking at the geometry. One of the nicest theorems I know is the following (see Hartshorne, Thm III.9.9):
Theorem. Let $T$ be an integral Noetherian scheme. Let $X \subseteq \mathbb P^n_T$ be a closed subscheme. For each point $t \in T$, we consider the Hilbert polynomial $P_t \in \mathbb Q[z]$ of the fibre $X_t$, considered as a closed subscheme of $\mathbb P^n_{\kappa(t)}$. Then $X$ is flat over $T$ if and only if the Hilbert polynomial $P_t$ is independent of $t$.
What this says is that not only the dimension of a finitely generated module is constant in the fibres, but more generally if we have a projective morphism (a finite ring morphism is a special case of this), then the entire Hilbert polynomial is constant if and only if the morphism is flat.
You don't necessarily have to know precisely what this means to appreciate it: the Hilbert polynomial captures numerical invariants like the (Krull) dimension and the degree (for a finite morphism, this is the (linear) dimension of the module fibre), so for a flat family the fibre dimensions cannot jump. Often when you want to prove something is not flat, you have some geometric reason for it.
As an example of this, consider the ring homomorphism
\begin{align*}
\phi \colon k[x,z] &\to k[x,y]\\
x &\mapsto x,\\
z &\mapsto xy.
\end{align*}
Maximal ideals of $k[x,z]$ are of the form $\mathfrak m = (x-a, z-b)$, and for $a \neq 0$, $x$ becomes invertible in $B/\mathfrak mB$, so
$$B/\mathfrak mB = k[x,y]/(x-a,xy-b) = k[x,y]/(x-a,y-\tfrac{b}{a}) = k,$$
so the fibre is just a point (Spec of a field). On the other hand, when $a = 0$ and $b = 0$, we get
$$B/\mathfrak mB = k[x,y]/(x,xy) = k[x,y]/(x) = k[y],$$
so the fibre is one-dimensional. Geometrically, this is saying that the map
\begin{align*}
k^2 &\to k^2\\
(x,y) &\mapsto (x,xy)
\end{align*}
collapses the $y$-axis onto the origin, whereas everywhere else it is an isomorphism.
Thus, we conclude that $\phi$ is not flat, since some of its fibres have (Krull) dimension $1$, whereas most of them have dimension $0$.
