You have to be careful with your wording. Saying "a span is finite" doesn't really mean anything. The span of a collection of vectors is the set of all finite linear combinations of those vectors.
Consider the vector space of all real polynomials $\mathcal{P}(\mathbb{R})$. It has a basis $\{x^n \mid n \in \mathbb{N}\cup\{0\}\}$ which has infinite cardinality, so $\mathcal{P}(\mathbb{R})$ is infinite dimensional. Any finite linear combination of these polynomials gives you an element of $\mathcal{P}(\mathbb{R})$. If you were to take a finite collection of the polynomials, say $\{x^{n_i} \mid i =1, \dots, k\}$, then their span would not contain any polynomial of degree more than $\max_i n_i$, so you would not obtain $\mathcal{P}(\mathbb{R})$.
You must note that $A$ is a vector in $\Bbb R^2$ and $A'$ is a vector in $\Bbb R^3$. That being said, the span of $A$ is not the same as the span of $A'$ because they live in different spaces!
However, there is some sense in which these spans are the "same." Consider that for any vector $v \in \operatorname{span}A$, $v = \lambda(a, a)$ for some $\lambda \in \Bbb R$. Similarly, for any vector $u \in \operatorname{span}A'$, $u = \mu(a, a, a)$ for some $\mu \in \Bbb R$. Both vectors, though they live in different spaces, effectively encode one piece of data, namely the number $\lambda a$ or $\mu a$.
More precisely, we say there is an isomorphism, or a relabeling of these spaces with the set of real numbers, $\Bbb R$. That is, for any real number $r$, there is a vector in $\operatorname{span} A$ and $\operatorname{span} A'$ such that the components of that vector are each the number $r$. Similarly, given a vector in $\operatorname{span} A$ or $\operatorname{span} A'$, we can find a real number that is equal to that vector's single piece of data.
It is not the case that the span of any vector in $\Bbb R^2$ or $\Bbb R^3$ is equal to $\Bbb R$, but it is the case that they can be isomorphic to $\Bbb R$.
In symbols, we write
$$
\operatorname{span}A \cong \operatorname{span}A' \cong \Bbb R.
$$
To address your main question, the definition of a vector space is a set closed under an addition $+$ and scalar multiplication $\cdot$ such that various other conditions hold. The glaring issue with your saying that $\operatorname{span}A$ and $\operatorname{span}A'$ are equal is that we cannot add vectors in $\operatorname{span}A$ and $\operatorname{span}A'$, so this does not respect closure under addition.
Since not both of your premises are true, the conclusion that $\operatorname{span}A$ and $\operatorname{span}A'$ are equal is invalid. The conclusion that they can be isomorphic is valid, as I have described.
Best Answer
With your definition of dimension, almost any definable map you try will be a counterexample.
For example, let $V$ be a $1$-dimensional vector space. Then any set $X\subseteq V$ has dimension $1$, except for $\emptyset$ and $\{0\}$ (which both have dimension $0$).
Let $X = Y = V\setminus \{0\}$, and let $f\colon X\to Y$ be the identity map. Then for all $y\in Y$, $\dim(f^{-1}(y)) = \dim(\{y\}) = 1$. But $\dim(X) = 1$, while $\dim(Y) +1 = 2$.
Here is an alternative (and more sensible) definition of dimension for definable sets: Let $X\subseteq V^n$ be a definable set, and let's assume $V$ is $\omega$-saturated (if not, take a suitable elementary extension, and evaluate $X$ there). An element of $X$ is an $n$-tuple $a$, and we define $\dim(a)$ to be the size of a maximal independent sub-tuple of $a$, equivalently the dimension of $\text{span}(a)$ in your sense. Then we define $\dim(X)$ to be $\max_{a\in X} \dim(a)$, the maximal dimension of an element of $X$.
This definition of dimension agrees with the Morley rank relative to the theory of infinite $K$-vector spaces, and it satisfies the additivity condition in your question.
In the comments, you asked some follow-up questions.
No - as I said in my answer, my definition of dimension coincides with Morley rank.
There's a confusing thing about strongly minimal theories, which is that there are two reasonable meanings of "dimension", which are very different. Let $T$ be a strongly minimal theory.
Let $M\models T$, and let $A\subseteq M$ (not necessarily definable!). Define $\dim_1(A)$ to be the cardinality of a basis for $\text{acl}(A)$.
Let $X$ be a definable set, defined by $\varphi(x)$. Let $\mathbb{M}\models T$ be a monster model of $T$ (or at least $\omega$-saturated). Define $\dim_2(X)$ to be the maximal value of $\dim_1(a)$ over all tuples $a\in X(\mathbb{M})$.
The fact that $\dim_2$ coincides with Morley rank is Theorem 6.2.19, together with Lemma 6.2.16(i), in Marker.
In vector spaces, $\dim_1$ coincides with the ordinary notion of linear dimension, for closed sets. But it's unsatisfactory for definable sets, since it isn't invariant under elementary extensions. For example, let $T$ be the theory of $\mathbb{R}$-vector spaces, and consider the formula $\phi(x): x = x$. We have $\dim_1(\varphi(\mathbb{R}^2)) = 2$, but $\dim_1(\varphi(\mathbb{R}^{42})) = 42$, despite the fact that $\mathbb{R}^2\preceq \mathbb{R}^{42}$. So $\dim_1$ is really assigning a dimension to a particular subset of a particular model.
If you're familiar with field theory or algebraic geometry, it might be morehelpful to think about the theory of algebraically closed fields, where both notions of dimension are familiar mathematical notions.
If $A\subseteq K\models \text{ACF}_0$, then $\dim_1(A)$ is the transcendence degree of the field $\overline{\mathbb{Q}(A)}$ over $\mathbb{Q}$.
If $X$ is a definable set (which is a Boolean combination of zero-sets of polynomials), then $\dim_2(X)$ is the Krull dimension of the Zariski closure of $X$.
Yes, in any infinite vector space $V$, $\text{acl}(X) = \text{span}(X)$ for all sets $X$. The inclusion $\text{span}(X)\subseteq \text{acl}(X)$ is clear, and for the other, note that if $b\notin \text{span}(X)$, then $b$ has infinitely many images under automorphisms of $V$, so it's not in $\text{acl}(X)$.
Of course, if $V$ is finite (which can only happen if the base field is finite), then $\text{acl}(X) = V$ for any $X$.
Yes, it does. Basically, this comes down to the fact that in any strongly minimal structure, for any tuples $a$ and $b$, we have $$\dim(ab) = \dim(a/b) + \dim(b).$$ Here $\dim$ is what I wrote as $\dim_1$ above. For a proof, see here.