Rules:

This is a mashup of nonogram and 0h n0.
Every cell in the grid must be shaded either blue or red. The clues on the outside give the number of cells in each consecutive run of red cells, separated by at least one blue cell. The clues in the grid are always shaded blue, and give the total number of blue cells that can be seen in each of the four cardinal directions before being blocked by a red cell.
A circle in the grid has the same value as the associated circle in that column’s nonogram clue. Likewise, a square in the grid has the same value as the associated square in that row’s nonogram clue.

Solution:

Name of puzzle:

This puzzle is a mashup of nonogram and 0h n0, so it would fittingly be called 0h n0n0gram!

Rules:

This is a mashup of nonogram and 0h n0.
Every cell in the grid must be shaded either blue or red. The clues on the outside give the number of cells in each consecutive run of red cells, separated by at least one blue cell. The clues in the grid are always shaded blue, and give the total number of blue cells that can be seen in each of the four cardinal directions before being blocked by a red cell.
A circle in the grid has the same value as the associated circle in that column’s nonogram clue. Likewise, a square in the grid has the same value as the associated square in that row’s nonogram clue.

Solution:

Name of puzzle:

[Thanks to Stiv for aha/image!]
If we look at each 3*3 block of the solved grid, the blue cells spell out THIS IS AN OHNOGRAM!

Solve path:

We can begin with some nonogram deductions, treating the shapes purely as unknown, non-zero numbers.

By 0h n0 rules, the C4 circle must have a value of 2. We can then complete the column.

0h n0 rules say that the C5 circle must be at least 2, and nonogram rules say that it can be at most 2, so it is exactly 2. Some nonogram logic follows.

0h n0 says the C3 circle must be at least 4, which places one red cell in the column with nonogram. Nonogram then says the R5 square must be 2, so we can place a couple of cells using 0h n0, and another using nonogram. Now 0h n0 says the C3 circle must be 5, and we can finish the column.

0h n0 says the C1 circle must be 5, and we can complete the column. We can also trivially say that the R11 square must be 2, which doesn't give any extra cells.

Nonogram says the R1 square must be 3, and we can place a few cells from there. Then, 0h n0 says the R2 square must be 2, and nonogram gives even more cells.

0h n0 says the bottom C7 circle must be 1, and nonogram places cells. From there, 0h n0 says the top C7 circle must be 4, and we finish the row. Nonogram logic follows.

C6 circle and C9 circle can both be shown to be 3 by 0h n0, and C11 circle can be shown as 2 by 0h n0 and nonogram. Nonogram finishes the grid, and we can trivially fill in the rest of the numbers (C8 circle and R7 square both as 2).