Looks like the correct approach. Note that what you're showing is that
$$V \times W = \{(x_1, \dots,x_n,y_1, \dots, y_m) \in k^{n+m} \mid f_1(x) = \dots = f_s(x) = 0, g_1(y) = \dots = g_t(y) = 0 \}.$$
If you feel uncomfortable because the defining equations should be polynomials in $x, y$ (instead of just in $x$ for the $f$'s and just in $y$ for the $g$'s), you might want to make things explicit.
Say $\hat {f_i}(x_1,\dots,x_n,y_1,\dots,y_m) = f_i(x_1,\dots,x_n)$ (i.e., $f_i$ explicitly considered as a polynomial in $x, y$) and similarly $\hat{g_j}(x_1,\dots,x_n,y_1,\dots,y_m) = g_j(y_1,\dots,y_m)$. Phrased this way, you're showing that
$$V \times W = \{(x_1, \dots,x_n,y_1, \dots, y_m) \in k^{n+m} \mid \hat{f_1}(x,y) = \dots = \hat{f_s}(x,y) = 0, \hat{g_1}(x,y) = \dots = \hat{g_t}(x, y) = 0 \},$$
but the right-hand side of this is of course still equal to
$$\{(x_1, \dots,x_n,y_1, \dots, y_m) \in k^{n+m} \mid f_1(x) = \dots = f_s(x) = 0, g_1(y) = \dots = g_t(y) = 0 \},$$
so that is really just a notational issue.
It is true that every algebraic set is a finite union of algebraic varieties (irreducible algebraic sets), and this union is unique up to reordering. These irreducible pieces of an algebraic set are called the irreducible components. This all follows from the fact that a polynomial ring over a field is Noetherian, so that an algebraic set with the Zariski topology is a Noetherian topological space.
As an example, I always think of the algebraic set defined by the ideal $(xz,yz),$ which is not prime. The real picture of this algebraic set is a line through a plane, and these two objects are exactly the irreducible components of the algebraic set. Here is the picture:
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The components are defined by the prime ideals $(z)$ and $(x,y)$ which are the two minimal prime ideals containing $(xz,yz)$. This may be the eyeball test you desire, as most people would look at this set and say it is made of two parts. In general, the irreducible components of an algebraic set defined by an ideal $I$ correspond exactly to the minimal prime ideals containing $I$.
Concerning your second question, it is not easy in general to determine when an ideal is prime. I asked a question here seeking different techniques to detect when ideals are prime. It is often easier to see that an ideal is not prime, as in the example I've given.
Best Answer
Let me upgrade my comment in to an answer. Your first definition is one of many for affine varieties. The second definition is for an "affine space", which is a vector space where we forget where $0$ is - that means we're allowed to add any two elements of our space as long as the coefficients sum to one, but we don't have scalar multiplication, a choice of a $0$ vector, etc.
One thing that may be contributing to your confusion is that the French word "variété" has multiple meanings - this has been discussed here on MSE in the past.