I wish to understand geometrically (not just algebraically) why the dimension of the Grassmanian $G(k,n)$ is $k(n-k)$ and the dimension of a flag manifold $F(k_{1},k_{2},...,k_{n},N)$ is $\sum_{i=1}^{n}k_{i}(k_{i-1}-k_{i})+Nk_{n}$ (in fact with understanding the Grassmanian case it would be enough because the flag are just "nested" Grassmanians).
I am thinking in a spatial way in the well known $G(2,5)$ but I am unable to see geometrically how the space of all $2$-planes in $\mathbb{P}^{5}$ can be $6$-dimensional...
Just like projective space $\mathbb{P}(V)$, the Grassmannian $G(k, V)$ has an covering by open sets, giving it its manifold structure. This covering is really the thing you should understand - then the dimension of the Grassmannian will just be the dimension of any of these open sets. The trick is to find a way of describing the affine patches of projective space which generalises in a nice way to the Grassmannian.
Let's suppose for the moment we are working with $V = \mathbb{R}^3$, so that $G(1, V)$ is the projective plane. Take a point $B \in G(1, V)$, then we want to find a nice open set $U_B$ containing $B$. There is an obvious choice: let $U_B \subseteq G(1, V)$ be those lines which are not contained in the plane $B^\perp$. Usually in projective geometry at this point, you write down coordinates for each line $L \in U_B$ by intersecting $L$ with some shift of the plane $B^\perp$, and hence ultimately end up giving an identification $U_B \cong B^\perp$.
There is something slightly more natural we can do, ending up with an identification $U_B \cong \operatorname{Hom}(B, B^\perp)$, the space of linear maps $B \to B^\perp$. Because of the direct sum decomposition $V = B \oplus B^\perp$, we have orthogonal projection maps $p: V \to B$ and $q: V \to B^\perp$. Given a line $L \in U_B$, the map $p|_L: L \to B$ will be an isomorphism (every line in $U_B$ has some projection onto $B$), and its inverse can be composed with $q|_L: L \to B^\perp$ to get a map $\varphi_L \in \operatorname{Hom}(B, B^\perp)$: $$ \varphi_L: B \to B^\perp, \quad B \xrightarrow{(p|_L)^{-1}} L \xrightarrow{q|_L} B^\perp$$ Conversely, given a map $\alpha: B \to B^\perp$, we may find the corresponding $L$ as the image of $b \mapsto b + \alpha(b)$. So we have an isomorphism $U_B \cong \operatorname{Hom}(B, B^\perp)$. (Note that if we fix some $b_0 \in B$, and just remember $\varphi_L(b_0) \in B^\perp$ for each line $L$, then we recover the usual view of projective coordinates as an intersection of the line and a shift of $B^\perp$). The map $\varphi_L$ is telling us how to "add some extra stuff" as we travel up the line $B$, in order to recover the more general line $L$.
So, after you've spent a bit of time drawing this to convince yourself it works, it takes almost no effort to generalise to Grassmannians. Let $V$ be an $n$-dimensional vector space, and $B \in G(k, V)$ a $k$-plane in $V$. Then define the open set $U_B \subseteq G(k, V)$ to be those $k$-planes complementary to $B^\perp$: $$ U_B = \{W \in G(k, V) \mid W \cap B^\perp = 0 \} $$ and again, using the restriction of the orthogonal projections $p: V \to B$ and $q: V \to B^\perp$ to a $k$-plane $W \in G(k, V)$ we may build a linear map $$ \varphi_W: B \to B^\perp, \quad B \xrightarrow{(p|_W)^{-1}} W \xrightarrow{q|_W} B^\perp$$ making $U_B \cong \operatorname{Hom}(B, B^\perp)$. The dimension of $\operatorname{Hom}(B, B^\perp)$ is $k(n-k)$, and this vector space is also naturally identified with the tangent space of $G(k, V)$ at the point $B$.
A last comment: note that if $V = \mathbb{R}^3$ again and $B \in G(2, V)$ is a $2$-plane, then $B^\perp$ is a line, and giving an element $\alpha \in \operatorname{Hom}(B, B^\perp)$ is tantamount to giving a "height function" over the plane $B$, again allowing us to lift up the reference plane $B$ to more general planes.