Return to Answer

Now note that the formation of Trace map is compatible with restrictions to open subsets. Hence, the composition $Tr_f\circ f^*:\Omega^q_{Y/k} \to \Omega^q_{Y/k}$ is the identity morphism on an open dense subset $U\subset Y$ (because $f$ is birational!). Since $\Omega^q_{Y/k}$ is a locally free sheaf, we conclude that $Tr_f\circ f^*$ is also the identity morphism. Therefore, $H^p(Tr_f\circ f^*)=H^p(Tr_f)\circ H^p(f^*)$ is the identity morphism as well. So $H^p(f^*)$ is injective as desired.

Now note that the formation of Trace map is compatible with restrictions to open subsets. Hence, the composition $Tr_f\circ f^*:\Omega^q_{Y/k} \to \Omega^q_{Y/k}$ is the identity morphism on an open dense subset $U\subset Y$ (because $f$ is birational and Y is normal!). Since $\Omega^q_{Y/k}$ is a locally free sheaf, we conclude that $Tr_f\circ f^*$ is also the identity morphism. Therefore, $H^p(Tr_f\circ f^*)=H^p(Tr_f)\circ H^p(f^*)$ is the identity morphism as well. So $H^p(f^*)$ is injective as desired.

Moreover, there is an example of a smooth compact non-Kahler complex manifold X, s.t. the Hodge-to-de Rham spectral sequence degenerates but the the Hodge filtration on de Rham cohomology $H^n(X,\mathbf C)$ doesn't define pure Hodge structure. In particular, hodgy symmetry fails for $h^{0,1}$ and $h^{1,0}$. An explicit situation of such a manifold is a Hopf surface $X=(\mathbf C^2 - \{0,0\})/q^{\mathbf Z}$, where $q\in \mathbf C^*$ and $|q|<1$ (action is diagonal). Then one can prove that Hodge-to-de Rham spectral sequence degenerates for $X$ but $H^1(X,\mathcal O_X)=0\neq \mathbf C=H^0(X,\Omega^1_{X/\mathbf C})$. Although, it doesn't imply that there is no canonical isomorphism $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^q(X,\Omega^p_{X/\mathbf C})$, it shows that it's unlikely that the 2nd statement can be a consequence of the first.

Therefore $\sum_{i\geq p}h^{i,n-i}=\sum_{i\leq n-p}h^{i,n-i}$! And we showed that this is equivalent to the equality $$ F^pH^n_{dR}(X)+\overline{F^{n-p+1}H^n_{dR}(X)}=H^n_{dR}(X). $$ In other words we showed that $(H^n(X,\mathbf Z), F^{\bullet}H^n(X,\mathbf C))$ is a pure Hodge structure of weight $d$.

UPD:

Moreover, there is an example of a smooth compact non-Kahler complex manifold X, s.t. the associated Hodge-to-de Rham spectral sequence degenerates but the the Hodge filtration on the de Rham cohomology $H^n_{dR}(X)$ doesn't define a pure Hodge structure. An explicit situation of such a manifold is a Hopf surface $X=(\mathbf C^2 - \{0,0\})/q^{\mathbf Z}$, where $q\in \mathbf C^*$ and $|q|<1$ (action is diagonal). Then one can prove that Hodge-to-de Rham spectral sequence degenerates for $X$ but $H^1(X,\mathcal O_X)=0\neq \mathbf C=H^0(X,\Omega^1_{X/\mathbf C})$. Hence the Hodge filtration doesn't define a pure Hodge structure on $H^1_{dR}(X)$ because of the failure of hodge symmetry. Although, it doesn't imply that there is no canonical isomorphism $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^q(X,\Omega^p_{X/\mathbf C})$, it shows that it's unlikely that the 2nd statement can be a consequence of the first.

Therefore $\sum_{i\geq p}h^{i,n-i}=\sum_{i\leq n-p}h^{i,n-i}$! And we showed that this is equivalent to the equality $$ F^pH^n_{dR}(X)+\overline{F^{n-p+1}H^n_{dR}(X)}=H^n_{dR}(X). $$ In other words we showed that $(H^n(X,\mathbf Z), F^{\bullet}H^n(X,\mathbf C))$ is a pure Hodge structure of weight $d$.

Let $X$ be a smooth algebraic variety over $\mathbf C$. Then one can define a descending filtration on the algebraic de Rham complex $\Omega^{\bullet}_{X/\mathbf C}$ as $F^i\Omega^{\bullet}_{X/\mathbf C}=\Omega_{X/\mathbf C}^{\geq i}$. This induces a descending filtration $F^{\bullet}H^n_{dR}(X)=F^{\bullet}H^n(X,\mathbf C)$. The main result is that this construction defines a pure Hodge structure under the properness assumption on X.

Therefore $\sum_{i\geq p}h^{i,n-i}=\sum_{i\leq n-p}h^{i,n-i}$! And we showed that this is equivalent to the equality $$ F^pH^n_{dR}(X)+\overline{F^{n-p+1}H^n_{dR}(X)}=H^n_{dR}(X). $$ In other words we showed that $(H^n(X,\mathbf Z), F^{\bullet}H^n(X,\mathbf C))$ is a pure Hodge structure of weight $d$.

First of all, this is true that Hodge decomposition holds for smooth proper varieties over $\mathbf C$. However, the standard proof (that is explained below) uses the projective case as a black box.

Secondly, degeneration of Hodge-to-de Rham spectral sequence does imply that there is a filtration on $H^{n}(X,\mathbf C)$ s.t. the associated graded pieces are isomorphic to $H^{p,q}(X)$. In particular, it means that there is an abstract isomorphism of $\mathbf C$ vector spaces $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^q(X,\Omega^p_{X/\mathbf C})$. But I don't know any way to pick a canonical one just assuming degeneration of the spectral sequence.

Moreover, there is an example of a smooth compact non-Kahler complex manifold X, s.t. the Hodge-to-de Rham spectral sequence degenerates but the the Hodge filtration on de Rham cohomology $H^n(X,\mathbf C)$ doesn't define pure Hodge structure. In particular, hodgy symmetry fails for $h^{0,1}$ and $h^{1,0}$. An explicit situation of such a manifold is a Hopf surface $X=(\mathbf C^2 - \{0,0\})/q^{\mathbf Z}$, where $q\in \mathbf C^*$ and $|q|<1$ (action is diagonal). Then one can prove that Hodge-to-de Rham spectral sequence degenerates for $X$ but $H^1(X,\mathcal O_X)=0\neq \mathbf C=H^0(X,\Omega^1_{X/\mathbf C})$. Although, it doesn't imply that there is no canonical isomorphism $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^q(X,\Omega^p_{X/\mathbf C})$, it shows that it's unlikely that the 2nd statement can be a consequence of the first.

All of that being said, I will explain how to construct pure Hodge structures on cohomology of any proper smooth variety.

Let us start with $X$ being a smooth algebraic variety over $\mathbf C$. Then one can define a descending filtration on the algebraic de Rham complex $\Omega^{\bullet}_{X/\mathbf C}$ as $F^i\Omega^{\bullet}_{X/\mathbf C}=\Omega_{X/\mathbf C}^{\geq i}$. This induces a descending filtration $F^{\bullet}H^n_{dR}(X)=F^{\bullet}H^n(X,\mathbf C)$. The main result is that this construction defines a pure Hodge structure under the properness assumption on X.

Therefore $\sum_{i\geq p}h^{i,n-i}=\sum_{i\leq n-p}h^{i,n-i}$! And we showed that this is equivalent to the equality $$ F^pH^n_{dR}(X)+\overline{F^{n-p+1}H^n_{dR}(X)}=H^n_{dR}(X). $$ In other words we showed that $(H^n(X,\mathbf Z), F^{\bullet}H^n(X,\mathbf C))$ is a pure Hodge structure of weight $d$.

UPD: