I will assume varieties are irreducible. Let $I$ be the kernel of $f^\sharp : \mathbb{C}[W] \to \mathbb{C}[V]$. Since $f^\sharp$ is surjective, $I$ must be a prime ideal, so corresponds to some closed subvariety $Y \subseteq W$. Thus the morphism $f : V \to W$ must factor through the inclusion $Y \hookrightarrow W$. We may assume without loss of generality that $Y = W$. But then $f^\sharp$ is an isomorphism, say with two-sided inverse $g^\sharp : \mathbb{C}[V] \to \mathbb{C}[W]$, and the fundamental theorem regarding morphisms of varieties implies the morphism $g : W \to V$ corresponding to $g^\sharp$ must be a two-sided inverse for $f : V \to W$.
You are right that the image of a morphism of affine varieties need not be closed: for example, if $V = \{ (x, y) \in \mathbb{C}^2 : x y = 1 \}$, $W = \mathbb{C}$, and $f(x, y) = x$, then the image of $f$ is a not closed in $W$. But what about $f^\sharp$? Well, $\mathbb{C}[V] = \mathbb{C}[x, y] / (x y - 1)$ and $\mathbb{C}[W] = \mathbb{C}[x]$, so $f^\sharp : \mathbb{C}[V] \to \mathbb{C}[W]$ is not surjective. (In fact, it is injective!) Thus the argument in the previous paragraph does not apply.
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
I'll treat the case $\dim V = 1$. To show a one-dimensional variety of degree $1$ is a line, we show that it is contained in any hyperplane passing through two points on $V$. So choose $x,y \in V$ with $x\neq y$ and let $H$ be a hyperplane containing $x$ and $y$. Notice $H$ is a linear variety of codimension $1$.
If $V \nsubseteq H$, the projective dimension theorem implies that every irreducible component of $V \cap H$ has dimension $0$. But now because $V$ has degree $1$, this means that $V \cap H$ can only contain $1$ point, contradicting $x,y \in V \cap H$. It follows $V$ is contained in every hyperplane passing through any two points of it, so that $V$ is a line (exercise ).