Axler's Theorem 2.43: the dimension formula There are a few steps of Axler's proof of the dimension formula that I'm not certain I fully follow. The statement is:

If $U_1$ and $U_2$ are subspaces of a finite-dimensional vector space $V$, then
$$
\dim(U_1 + U_2) = \dim U_1 + \dim U_2 - \dim (U_1 \cap U_2). 
$$

I'm going to call the ambient vector space $V$.  I'm going to try to replicate the proof, pausing at the points I don't fully understand. I'm going to try to stick to Axler's notation because I'd like to quote from the text later on.

Define $m: = \dim U_1 \cap U_2$ and choose a basis $u_1, \ldots, u_m$ for $U_1 \cap U_2$. As $U_1 \cap U_2 \subset U_1$ and $U_1 \cap U_2 \subset U_2$, and linear independence in a subset implies linear dependence in the ambient set (in this case, $U_1$ and $U_2$), we can extend $u_1, \ldots, u_m$ to a basis of $U_1$ and $U_2$. Call the basis for $U_1$ $u_1, \ldots, u_m, v_1, \ldots, v_j$, and the basis for $U_2$ $u_1, \ldots, u_m, w_1, \ldots, w_k$. We'll show that the full concatenation of the $u$'s, $v$'s, and $w$'s is a basis for $U_1 + U_2$.

So far, I'm ok with this setup. It uses the facts that a linearly independent set can be extended to a basis, that $U_1 \cap U_2$ is a subspace of both $U_1$ and $U_2$, and that linear independence in $U_1 \cap U_2$ implies linear independence in $U_1$ and $U_2$, and the $u_i$ are linearly independent because they form a basis for $U_1 \cap U_2$.
Continuing with the proof, the next step is what caused me for the most trouble, and I'm going to quote Axler directly.

"Clearly $\mathrm{span}(u_1, \ldots, u_m, v_1, \ldots, v_j, w_1, \ldots, w_k)$ contains $U_1$ and $U_2$ and hence equals $U_1 + U_2$. So to show that this list is a basis of $U_1 + U_2$ we need only show that it is linearly independent."

(I'm going to define this span as $S$, for ease of writing.)
It makes sense to me that $U_1$ and $U_2$ are contained in the span. I can write any $u \in U_1$ in terms of the $u_i$ (they're a basis), and then just set the coefficients of the $v$'s and $w$'s to $0$. The procedure for showing containment of $U_2$ is similar, setting the coefficients of the $v$'s to $0$ and using the fact that the $u$'s and $w$'s form a basis for $U_2$.
It's also true, and was proved earlier, that $U_1 + U_2$ is the smallest subspace containing $U_1$ and $U_2$. So since $S$ contains $U_1$ and $U_2$, we have $U_1 + U_2 \subset S$. Axler doesn't seem to prove the other direction, might it just be taken as obvious. This spanning list contains a basis for $u_1$ and for $u_2$, so I can certainly use to write any element of $U_1 + U_2$. But I don't think we even need equality. If $S \supset U_1 + U_2$, I believe we are able to say that $S$ spans $U_1 + U_2$ (I'm not sure if we need equality to say that, but it follows from the above; this may just be a matter of different definitions). So it suffices to show linear independence since we already have spanning.
I'd appreciate some help on parsing this proof.
 A: The point you seem to be hung up on is understanding why $\dim(U_1 + U_2) = m + n + o$ when we let $\dim(U_1) = m+n$ and $\dim(U_2) = m + o$ and $\dim(U_1 \cap U_2) = m$ such that the stated formula makes sense.
For a list of vectors to be a basis, they must span the space and they must be linearly independent in the space. If we can show that the list $u_1, ...,u_m,v_1,...v_n,w_1,...,w_o$ has both of these properties then we are finished (I am assuming you are fine with the step where the basis of $U_1 \cap U_2$ was extended to form basises of the spaces $U_1$ and $U_2$.
Then, we can show this list spans rather easily by considering that by definition some $U_1 + U_2$ is given by the sum of all $u \in U_1$ and any $w \in W$ . Therefore, $v \in U_1 + U_2 = u + w =(a_1u_1+...+a_mu_m+b_1v_1+...+b_nv_n) + (c_1u_1 + ... + c_mu_m + d_1w_1 + ... + d_ow_o)$
Which follows by considering the representation of $u$ and $w$ as a linear combination of the basis vectors of the space they are within. This can be rearranged to get a linear combination of the list we are trying to prove spans the space and therefore this list does span the space.
The less obvious part of this proof is proving that this list is linearly independent in the space. However, noting what we know will be useful in reaching the conclusion independently. That is, we know that the list of vectors $u_1, ... u_m$ is linearly independent, that the list of vectors $u_1, ..., u_m, v_1, ..., v_n$ is linearly independent, and that the list $u_1, ... , u_m, w_1,.., w_o$ is linearly independent. (all implicit in the fact that these lists form a basis.)
To show it, we must show that the only linear combination of the vectors $u_1, ..., u_m, v_1, ..., v_n, w_1, ..., w_o$ that is equal to 0 is the one where all of the coefficients are equal to zero. If we consider some vector $v \in U_1 + U_2$ we can write it in the form $v = a_1u_1 + ... + a_mu_m + b_1v_1 + ... + b_nv_n + c_1w_1 + ... + c_ow_o = 0$, and rearrange to obtain $c_1w_1 + ... + c_ow_o = -a_1u_1 - ... - a_mu_m - b_1v_1 - ... - b_nv_n$.
By definition, this means that $c_1w_1 + ... + c_ow_o \in U_1$, whatever we choose $c_1, c_2, ..., c_o$. Which means that $v$ can be written entirely in terms of the vectors $u_1, ... u_m, v_1, ... v_n$, assuming that we are saying it equals to zero.. But in this case, we can see that this list has already been shown to be linearly independent. And therefore the only way to let $v = 0$ is by setting all coefficients equal to zero, by the linear independence of the list $u_1, ... u_m, v_1 ... v_n$.
Understanding that the fact that $v \in V$ means that v is equal to some linear combination of the basis vectors of $V$ is the most important part of understanding this proof.
