I know that Bezout's theorem says that if you take two plane curves, then their maximal number of intersection points is the product of their degrees. However, assume that I have two irreducible polynomials in say 6 variables and I plug in values for 4 of them. I'm left with two plane curves. What can be said about their number of intersections?
I assume the original two polynomials are irreducible. I know that there are cases like
$$u+f(x,y),\;\; v+f(x,y),$$
where I can plug in zero for $u,v$ and then I end up with just one curve. Therefore, is it possible to say something like ''for most values of the other 3 variables'' the number of intersection points is the product of the degrees?
Yes, what you want to say is true.
Let's stick to projective space and homogeneous equations to keep things tidy. (There's a process called "homogenisation" that allows us to pass from inhomogeneous polynomials in $n$ variables to homogeneous polynomials in $n+1$ variables of the same degree which then define hypersurfaces in projective space $\mathbf{P}^n.$ So we can answer the projective version first, and then go back to the original version.)
Suppose your two original polynomials homogenise to give $F$ and $G$. Then they define two hypersurfaces $\{F=0\}$ and $\{G=0\}$ in projective space $\mathbf P^6$. Since $F$ and $G$ are irreducible and distinct, these two hypersurfaces intersect in a subset $X$ of dimension 4 and degree $(\operatorname{deg} F \cdot \operatorname{deg} G)$.
Now "plugging in values" for 4 variables in your setting corresponds, in the projective setting, to intersecting with a linear subspace $L$ of codimension 4, in other words a plane.
Then the statement you want is that for "almost all" choices of the plane $L$, the intersection $X \cap L$ will consist of $(\operatorname{deg} F \cdot \operatorname{deg} G)$ points in $\mathbf P^6$. Of course, there are some choices of $L$ for which this will fail --- it could be that $\{F=0\} \cap \{G=0\}$ contains a line, for example, in which case any plane $L$ containing that line will intersect in an infinite set of points. But those "bad" $L$ are a smaller-dimensional set of the set of all possible choices of $L$.
Of course you asked about inhomogeneous polynomials, so let's go back to the affine picture. For a slightly smaller set of choices of $L$ than in the previous paragraph, none of the points of $X \cap L$ will be "points at infinity" in $\mathbf P^6$, so your original equations have exactly $(\operatorname{deg} F \cdot \operatorname{deg} G)$ common solutions.