Andrew's answer is correct, but here is a more geometric way to think about the same answer.
If a codimension $r$ variety is cut out by $r$ equations, then each equation
genuinely cuts out a new dimension, i.e. $V(f_1,\ldots,f_{i+1})$ is a divisor
in $V(f_1,\ldots,f_i)$ cut out by $f_{i+1} = 0$. Thus the intersection
of $V(f_1,\ldots,f_i)$ and $V(f_{i+1})$ is a proper intersection, and so the
degrees multiply.
Here is
one way to think about this last statement:
We can assume that the degree of $V(f_1,\ldots,f_i)$ is equal to $d,$ the product of the degrees of $f_1, \ldots,f_i$, by induction. Geometrically, this means that a generic linear subspace $L$ of dimension $i$ meets $V(f_1,\ldots,f_i)$ in $d$ points.
To compute the degree of $V(f_1,\ldots,f_{i+1})$ we intersect with a generic linear subspace $L'$ of dimension $i+1$.
Suppose that $d'$ is the degree of $f_{i+1}$. Then you can deform the equation $f_{i+1} = 0$ to an equation of the form $l_1\ldots l_{d'}$, where each $l_j$ is a generically chosen linear equation.
Thus $V(f_{i+1})$ can be deformed to the union of the hyperplanes $V(l_1)\cup \cdots \cup V(l_{d'})$, and so
$$V(f_1,\ldots,f_{i+1}) \cap L' = V(f_1,\ldots,f_i) \cap V(f_{i+1}) \cap L',$$ which can be deformed to
$$V(f_1,\ldots,f_i) \cap \bigl(V(l_1)\cup \cdots \cup V(l_{d'})\bigr) \cap L'
= \bigl(V(f_1,\ldots , f_i) \cap V(l_1) \cap L'\bigr)
\cup \cdots \cup
\bigl(V(f_1,\ldots,f_i)\cap V(l_1)\cap L'\bigr) \bigr).$$
Now each $V(l_j)\cap L'$ is the intersection of a generic hyperplane and the generic $i+1$-dimensional linear subspace $L'$, and so is a generic linear subspace of dimension $i$. Thus each intersection $V(f_1,\ldots,f_i)\cap V(l_j)\cap L'$ consists of $d$ points, and so their union consists of $dd'$ points. QED
If $V$ is an $r$-codimensional variety cut out by more than $r$ equations, then
we can write it in the form $V(f_1,\ldots,f_r, f_{r+1},\ldots,f_s)$ for some $s > r$, where $V(f_1,\ldots,f_r)$ is already of codimension $r$ (but not irreducible). The closed subscheme $V(f_1,\ldots,f_r)$ of $\mathbb P^n$ is then of degree equal to the product of the degrees of the $f_1,\ldots,f_r$, but (by assumption) the additional equations $f_{r+1}, \ldots, f_s$ do not cut down the dimension any further. Instead, they cut out some particular irreducible component of $V(f_1,\ldots,f_r)$. In particular, the intersections
$V(f_1,\ldots,f_r) \cap V(f_j)$ for $j > r$ are not proper, and the degree of a non-proper intersection is not given as the product of the degrees.
So the basic phenomenon here is as follows: if $V$ is a reducible algebraic set of (equi)codimension $r$, a union of components $V_1,\ldots,V_m$, then
$\deg V = \sum_i \deg V_i$, but there is no immediate relationship between
the degrees of the additional polynomials needed to cut out the various $V_i$
and the degrees of those $V_i$.
A good example to think about is the case of a twisted cubic curve in $\mathbb P^3$ (with hom. coords. $X,Y,Z,W$). It is obtained by first intersecting the quadrics $V(X^2 - YW)$ and $V(XZ - Y^2)$. This intersection is a degree $4$ curve which is reducible; it is the union of a line $L$ (the line cut out by $X = Y = 0$) and the twisted cubic curve $C$. To cut out $C$ we have to impose the additional equation $X^3 - ZW^2 = 0$.
As mentioned in the comments, the best way to prove statements like this is to use the Hilbert polynomial definition of degree. For this point of view, see section I.7 of Algebraic Geometry by Hartshorne.
Let $R = k[x_0,\ldots, x_n]$ where $k$ is an algebraically closed field viewed as a graded ring. For any graded $R$-module $M$, the Hilbert function is the function
$$
h(l) = \dim_k M_l
$$
giving the dimension as a $k$-vector space of the graded pieces of $M$. The idea is that for large enough $l$ this agrees with a polynomial $P_M(l)$ which is the Hilbert polynomial of $M$. Then for a projective variety $X \hookrightarrow \mathbb{P}^n$, the Hilbert polynomial $P_X(l)$ is just the Hilbert polynomial of the homogeneous coordinate ring of $X$ as a graded module over $R$. You can show that the degree of $P_X(l)$ is $d = \dim X$ and then we define the degree of $X$ to be $d!$ times the leading coefficient of $P_X(l)$
The nice thing about the Hilbert polynomial is that it behaves well with exact sequences and this gives it the geometric properties we want and expect. This is because $\dim_k$ is additive on exact sequences.
In particular, if we have that $X = Y_1 \cup Y_2$ with $Y_1$ and $Y_2$ the same dimension and intersecting in a lower dimension, then we can write the exact sequence
$$
0 \to R/I \to R/I_1 \oplus R/I_2 \to R/(I_1 + I_2) \to 0
$$
where $I_i$ is the homogeneous ideal of $Y_i$ and $I$ is the homogeneous ideal of $X$. Then by additivity of the Hilbert polynomial,
$$
P_{R/I_1 \oplus R/I_2} = P_{R/I} + P_{R/(I_1 + I_2)}.
$$
Applying additivity of the Hilbert polynomial again, we see that the left hand side of this equation is in fact $P_{R/I_1} + P_{R/I_2}$. Rephrasing this geometrically, we see that
$$
P_{Y_1} + P_{Y_2} = P_{X} + P_{Y_1 \cap Y_2}.
$$
Since $Y_i$ were assumed to be the same dimension, the leading coefficient of the left hand side is the sum $\deg{Y_1}/d! + \deg{Y_2}/d!$. Similarly, on the right hand side, since we assumed $Y_1 \cap Y_2$ is lower dimensional than all of $X$, we have that the leading coefficient of the right hand side is just that of $P_{X}$, that is, $\deg{X}/d!$, giving us the equality $\deg{X} = \deg{Y_1} + \deg{Y_2}$.
Now you can deduce the more general case with a little more work by applying this to the irreducible components.
The proof that this gives the same definition of degree as the one you gave is a little involved but it uses exactly the same technique. Write down an exact sequence whose terms correspond to the varieties we are intersecting and compare the two sides of the equation we get for the Hilbert polynomials. However, it requires some commutative algebra.
Edit: I wanted to add a bit about the dimension considerations since you brought that up as something you had issue with. If you notice, my argument above implies something a little different from your statement. It says that the degree of a variety is the sum of degrees of the highest dimensional irreducible components. This is because the contribution of the lower dimensional components to the Hilbert polynomial will not affect the leading coefficient which is the same degree as the dimension.
How does this reconcile with the classical notion of degree? The idea is that the lower dimensional components won't affect the intersection with your general plane. The reason for this is that if we have a $k$ dimensional subvariety $Y$ of $\mathbb{P}^n$, the classical degree is the number of points in the intersection with an $n - k$ plane. "Most" $n-k$ planes will certainly miss any components of dimension less than $k$ (think for example a point and a line in $\mathbb{P}^3$) and thus shouldn't contribute to the classical notion of degree, and indeed with the Hilbert polynomial argument, we see that they don't. Hopefully this fixes part of your confusion about how the dimension affects things.
Best Answer