Prove $\dim_F(U+V)+\dim_F(U\cap V)=\dim_F(U)+\dim_F(V)$ My note here is a bit floppy so I want to recover it.
Theorem:
$\dim_F(U+V)+\dim_F(U\cap V)=\dim_F(U)+\dim_F(V)$
Proof: [format: note::my comment]
Consider $f:U\bigoplus V\rightarrow U+V$ given by $f(u,v)=u+v$
Image of $f$ is $U+V$ :: I think this is something trivial instad of something to be proved. Saying $f$ is surjective. Move on.
$Ker(f)=\{(u,v)|u+v=0\}=\{(u,v)|u\in U,v\in V,u=-v\}=\{(u,-u)|u\in U\cap V\}$ :: (typo corrected, thanks for pointing out) I think this line is saying $Ker(f)=U\cap V$? But later I revoked this idea because then there's not point of proving isomorphism. We'll see.

If $u\in U\cap V$ then $u\in U$ then $u\in U$, $-u\in V$ and $u+(-u)=0$ :: Update: What is the point of putting this line here? Why is this needed for proving isomorphism?
Consider if $u+v=0$ then $u=-v\in U\cap V$ :: Update: What is the point of putting this line here? Why is this needed for proving isomorphism?

But the map $U\cap V\rightarrow Ker(f)$ is linear, bijective so $Ker(f)\simeq U\cap V$
But $\dim_F(Im(f))+\dim_F(Ker(f))=\dim_F(U\bigoplus V)$
i.e. $\dim_F(U+V)+\dim_F(U\cap V)=\dim_F(U)+\dim_F(V)$
Update:
 A: First comment: I do hope that you could write up the proof that the image of $f$ is $U + V$, because it is very easy.
Second comment: Yes, there is a typo. The working should be:
\begin{align*}
\operatorname{Ker}(f) &= \lbrace (u, v) | u \in U, v \in V, u + v = 0 \rbrace \\
&= \lbrace (u, v) | u \in U, v \in V, u = -v \rbrace \\
&= \lbrace (u, -u) | u \in U \cap V \rbrace.
\end{align*}
Also, tying into the fourth comment, no, this does not mean that $\operatorname{Ker}(f) = U \cap V$. You have to remember that $(u, -u)$ and $u$ are different objects. They are, however, definitely related, by the isomorphism argument. The map
$$\operatorname{Ker}(f) \rightarrow U \cap V : (u, -u) \mapsto u$$
is linear and bijective, so while the sets are not equal per se, they are isomorphic, and thus have the same dimension. Therefore,
$$\operatorname{dim} (U \oplus V) = \operatorname{dim}(\operatorname{Im(f)}) + \dim(\operatorname{Ker}(f))$$
becomes
$$\operatorname{dim}(U) + \operatorname{dim}(V) = \operatorname{dim}(U + V) + \operatorname{dim}(U \cap V).$$
A: Your comments in yellow boxes, my responses below them.

Image of $f$ is $U+V$ :: I think this is something trivial instad of something to be proved. Saying $f$ is surjective. Move on.

If this  is trivial to you, well  done. It should be,  by definition of $U + V$.

$\ker f = U \cap V$ :: But later I revoked this idea because then there's not point of proving isomorphism. We'll see. I think this line is  saying $\ker  f = U \cap V$.

Yes, this line is saying something very similar, but not equality. $ker f$ is a member of $U \oplus V$, while $U \cap V$ is a subset of both $U$ and $V$, and therefore cannot be equal to $\ker f$. You may feel that I am  pointing  out something very trivial, but it's good to iron out the details in your thought process, regardless of how miniscule they are. 
Why  is one calculating the kernel? Indeed, one is using an isomorphism theorem here, namely that if $T : U \to V$ is a linear transformation (in finite dimensions), then $\frac{U}{\ker{T}} \cong \operatorname{im} T$.  So calculating the kernel is important for us.
The next few lines just formally  prove that $\ker f \cong U \cap V$. These lines are important, because you may have understood why the congruence happens, but it has to be formally proved, is it not? The congruence, however obvious it may seem from the statement $\ker f = \{(u,-u) | u \in U \cap V\}$, must be proved, and that  is the purpose of the lines to which you have replied  "I don't know what's the point of putting this line here." Indeed, it's not saying the same thing, because the above statement is an equality of sets,  and the the following statements give an isomorphism of vector spaces, a far stronger proposition.
The last line is the application of the theorem I mentioned earlier. The point of  arguing an isomorphism, is that isomorphic spaces have the same dimension, so we can equate the dimensions, giving us the result.
I am  sure you will still have doubts. Clarify them with me.
