do you know how we cunstruct a blow up? for your first question blow up of affine $A$ at the ideal $I$ is covered by $$Spec(A[\frac{x^{r_0}\cdot y^{s_0}}{x^{r_i}\cdot y^{s_i}},...,\frac{x^{r_n}\cdot y^{s_n}}{x^{r_i}\cdot y^{s_i}}]),0\le i\le n$$, now let $Q$ be a prime ideal defining a point with positive dimension at center of $v$ over $$Spec(\mathcal{O}_{P, V_0}[\frac{x^{r_0}\cdot y^{s_0}}{x^{r_i}\cdot y^{s_i}},...,\frac{x^{r_n}\cdot y^{s_n}}{x^{r_i}\cdot y^{s_i}}])$$ and because you have finitely many dominator it define a point with positive dimension of one $Spec(A[\frac{x^{r_0}\cdot y^{s_0}}{x^{r_i}\cdot y^{s_i}},...,\frac{x^{r_n}\cdot y^{s_n}}{x^{r_i}\cdot y^{s_i}}])$ where $U= Spec A$ is a neighborhood of $P$.

For your second question, I don't completely understand your statement but the idea should be something like this: let's $v(x)=p,v(y)=q$ because the valuation is discrete, after a rescaling we can assume $(p,q)=1$. now we know that $$v(\sum a_{i j} x^iy^j)\le\min_{a_{i j}\neq 0}\{ip+jq\}$$ by third property of valuations,in fact from the third property we get that if $$v(a)\not =v(b),v(a+b)=min(v(a),v(b))$$, so we only have to handle the case that $\forall i,j,ip+jq$ is a constant. so we can assume that in $$t=\Sigma a_{ij}x^iy^j$$ powers of $x$ are strictly increasing. we prove by induction on the number of terms of $t$ that $$v(\sum a_{i j} x^iy^j)=ip+jq$$.assume that $$v(\sum a_{i j} x^iy^j)<ip+jq$$ then by invariance you mentioned $$v(\sum a_{i j} x^iy^j)=v(\sum a_{i j} (\alpha x)^iy^j)$$ lets $n$ be the biggest power of $x$ appearing in $t$ and take $\alpha^n=-1$ (base is algebraicly closed so such an $\alpha$ exist) now $$\sum a_{i j} x^iy^j+\sum a_{i j} (\alpha x)^iy^j$$ has less terms than $t$ and it's valuation is less than $v(t)<ip+jq$ which is a contradiction by induction hypothesis.

do you know how we cunstruct a blow up? for your first question blow up of affine $A$ at the ideal $I$ is covered by $$Spec(A[\frac{x^{r_0}\cdot y^{s_0}}{x^{r_i}\cdot y^{s_i}},...,\frac{x^{r_n}\cdot y^{s_n}}{x^{r_i}\cdot y^{s_i}}]),0\le i\le n$$, now let $Q$ be a prime ideal defining a point with positive dimension at center of $v$ over $$Spec(\mathcal{O}_{P, V_0}[\frac{x^{r_0}\cdot y^{s_0}}{x^{r_i}\cdot y^{s_i}},...,\frac{x^{r_n}\cdot y^{s_n}}{x^{r_i}\cdot y^{s_i}}])$$ and because you have finitely many dominator it define a point with positive dimension of one $Spec(A[\frac{x^{r_0}\cdot y^{s_0}}{x^{r_i}\cdot y^{s_i}},...,\frac{x^{r_n}\cdot y^{s_n}}{x^{r_i}\cdot y^{s_i}}])$ where $U= Spec A$ is a neighborhood of $P$.

For your second question, I don't completely understand your statement but the idea should be something like this: let's $v(x)=p,v(y)=q$ because the valuation is discrete, after a rescaling we can assume $(p,q)=1$. now we know that $$v(\sum a_{i j} x^iy^j)\ge\min_{a_{i j}\neq 0}\{ip+jq\}$$ by third property of valuations,in fact from the third property we get that if $$v(a)\not =v(b),v(a+b)=min(v(a),v(b))$$, so we only have to handle the case that $\forall i,j,ip+jq$ is a constant. so we can assume that in $$t=\Sigma a_{ij}x^iy^j$$ powers of $x$ are strictly increasing. we prove by induction on the number of terms of $t$ that $$v(\sum a_{i j} x^iy^j)=ip+jq$$.assume that $$v(\sum a_{i j} x^iy^j)>ip+jq$$ then by invariance you mentioned $$v(\sum a_{i j} x^iy^j)=v(\sum a_{i j} (\alpha x)^iy^j)$$ lets $n$ be the biggest power of $x$ appearing in $t$ and take $\alpha^n=-1$ (base is algebraicly closed so such an $\alpha$ exist) now $$\sum a_{i j} x^iy^j+\sum a_{i j} (\alpha x)^iy^j$$ has less terms than $t$ and it's valuation is bigger than $v(t)>ip+jq$ which is a contradiction by induction hypothesis.